学习笔记之15道简单算法题
2024-05-09 07:06:16
15道简单算法题
http://www.cnblogs.com/hlxs/archive/2014/06/06/3772333.html
(●—●) | 剑指Offer_编程题_牛客网
http://www.nowcoder.com/ta/coding-interviews?page=1
source code
https://github.com/haotang923/interview/tree/master/15%20simple%20algorithm%20problems
1:合并排序，将两个已经排序的数组合并成一个数组，其中一个数组能容下两个数组的所有元素;
1 #include <iostream>
2 using namespace std;
3
4 class Solution
5 {
6 public:
7 void MergeArray(int a[],int alen,int b[],int blen)
8 {
9 int len = alen + blen - 1;
10 alen --;
11 blen --;
12
13 // 用归并排序的思想把两个数组合并
14 while (alen >= 0 && blen >= 0)
15 {
16 if (a[alen] > b[blen])
17 a[len --] = a[alen --];
18 else
19 a[len --] = b[blen --];
20 }
21
22 while (blen >= 0)
23 a[len --] = b[blen --];
24 }
25 };
26
27 int main ()
28 {
29 Solution testSolution;
30 int arrA[] = {1, 2, 4, 6, 8, 0, 0, 0, 0, 0};
31 int arrB[] = {1, 3, 5, 7, 9};
32
33 testSolution.MergeArray(arrA, 5, arrB, 5);
34
35 for (int i = 0; i < 10; i ++)
36 cout << arrA[i] << endl;
37
38 getchar();
39
40 return 0;
41 }View Code
2:合并两个单链表;
- C++11最好用nullptr初始化指针得到空指针，同时尽量避免使用NULL。
- https://msdn.microsoft.com/zh-cn/library/4ex65770.aspx
- 复习struct，链表也不熟悉了，用当前节点还是next节点的时候出了几次错。
- struct_百度百科
- http://baike.baidu.com/link?url=PxVQxXElF_rtYMqYhfNBzsjvTl9cG2PSKTqFZGV3IBAEHosImlkuaSMqCIxnShEqtDD4xA8gISTtP7rS7TENj_
- struct (C++)
- https://msdn.microsoft.com/zh-cn/library/64973255.aspx
1 #include <iostream>
2 using namespace std;
3
4 struct ListNode
5 {
6 int val;
7 ListNode *next;
8 ListNode(int x) : val(x), next(NULL) {}
9 };
10
11 class Solution
12 {
13 public:
14 ListNode* MergeList(ListNode* aListNode, ListNode* bListNode)
15 {
16 // NULL case
17 if (aListNode == NULL)
18 return bListNode;
19 if (bListNode == NULL)
20 return aListNode;
21
22 ListNode* resHead = NULL;
23
24 if (aListNode->val < bListNode->val) {
25 resHead = aListNode;
26 aListNode = aListNode->next;
27 } else {
28 resHead = bListNode;
29 bListNode = bListNode->next;
30 }
31
32 // Reserve the head point
33 ListNode* tmpHead = resHead;
34
35 // PrintListNode(tmpHead);
36
37 // Merge two LinkList
38 // Remember to check if the current node is NULL
39 // assign ListNode to next resHead->next
40 // move resHead to the next ptr
41 while (aListNode != NULL && bListNode != NULL)
42 {
43 if (aListNode->val < bListNode->val) {
44 resHead->next = aListNode;
45 aListNode = aListNode->next;
46 } else {
47 resHead->next= bListNode;
48 bListNode = bListNode->next;
49 }
50 resHead = resHead->next;
51 }
52
53 // PrintListNode(tmpHead);
54
55 if (aListNode != NULL)
56 resHead->next= aListNode;
57
58 if (bListNode != NULL)
59 resHead->next= bListNode;
60
61 // PrintListNode(tmpHead);
62
63 return tmpHead;
64 }
65
66 void PrintListNode(ListNode* inListNode)
67 {
68 ListNode* tmpLN = inListNode;
69
70 while (tmpLN != NULL)
71 {
72 cout << tmpLN->val << endl;
73 tmpLN = tmpLN->next;
74 }
75 cout << endl;
76 }
77 };
78
79 int main ()
80 {
81 Solution testSolution;
82 ListNode* aHead = new ListNode(1);
83 ListNode* cur = aHead;
84
85 for (int i = 2; i < 10; i += 2)
86 {
87 ListNode* tmpLN = new ListNode(i);
88 cur->next = tmpLN;
89 cur = tmpLN;
90 }
91
92 // testSolution.PrintListNode(aHead);
93
94 ListNode* bHead = new ListNode(1);
95 cur = bHead;
96
97 for (int i = 3; i < 10; i += 2)
98 {
99 ListNode* tmpLN = new ListNode(i);
100 cur->next = tmpLN;
101 cur = tmpLN;
102 }
103
104 // testSolution.PrintListNode(bHead);
105
106 ListNode* head = testSolution.MergeList(aHead, bHead);
107
108 // testSolution.PrintListNode(head);
109
110 while (head != NULL)
111 {
112 cout << head->val << endl;
113 head = head->next;
114 }
115
116 getchar();
117
118 return 0;
119 }View Code
3:倒序打印一个单链表;
- 递归
- 链表的题目还得多练，又弄错了打印head还是cur节点。。。
1 #include <iostream>
2 using namespace std;
3
4 struct ListNode
5 {
6 int val;
7 ListNode *next;
8 ListNode(int x) : val(x), next(NULL) {}
9 };
10
11 class Solution
12 {
13 public:
14 void PrintListNodeReversely(ListNode* inListNode)
15 {
16 if (inListNode != NULL)
17 {
18 PrintListNodeReversely(inListNode->next);
19 cout << inListNode->val << endl;
20 }
21 }
22 };
23
24 int main ()
25 {
26 Solution testSolution;
27 ListNode* head = new ListNode(0);
28 ListNode* cur = head;
29
30 for (int i = 1; i < 10; i ++)
31 {
32 ListNode* tmpLN = new ListNode(i);
33 cur->next = tmpLN;
34 cur = cur->next; // cur->next == tmpLN
35 }
36
37 // Print head node
38 testSolution.PrintListNodeReversely(head);
39
40 getchar();
41
42 return 0;
43 }View Code
4:给定一个单链表的头指针和一个指定节点的指针，在O(1)时间删除该节点;
- http://www.cnblogs.com/pegasus923/archive/2010/10/04/1841891.html
- 重温链表插入删除操作，注意原文代码删除头结点处有问题，函数对头指针只是值传递，如果要删除需要传入头指针的指针。
1 #include <iostream>
2 using namespace std;
3
4 struct ListNode
5 {
6 int val;
7 ListNode *next;
8 ListNode(int x) : val(x), next(NULL) {}
9 };
10
11 class Solution
12 {
13 public:
14 void PrintListNode(ListNode* inListNode)
15 {
16 if (inListNode != NULL) {
17 cout << inListNode->val << endl;
18 PrintListNode(inListNode->next);
19 } else
20 cout << endl;
21 }
22
23 void Test(ListNode* inListNode)
24 {
25 delete inListNode;
26 inListNode = NULL;
27 cout << inListNode << endl;
28
29 PrintListNode(inListNode);
30 }
31
32 void DeleteNode(ListNode* pInHead, ListNode* pToBeDeleted)
33 {
34 cout << "Entry void DeleteNode(ListNode* pInHead, ListNode* pToBeDeleted)\n" << endl;
35
36 // Check NULL always
37 if (pInHead == NULL || pToBeDeleted == NULL)
38 return;
39
40 // PrintListNode(pInHead);
41
42 // Delete non-tail node including head node
43 if (pToBeDeleted->next != NULL) {
44 ListNode* pNext = pToBeDeleted->next;
45 pToBeDeleted->val = pNext->val;
46 pToBeDeleted->next = pNext->next;
47
48 delete pNext;
49 pNext = NULL;
50 } else { // Delete tail
51 ListNode* pPre = pInHead;
52
53 while (pPre->next != pToBeDeleted && pPre != NULL)
54 pPre = pPre->next;
55
56 if (pPre == NULL)
57 return;
58
59 pPre->next = NULL;
60 delete pToBeDeleted;
61 pToBeDeleted = NULL;
62 }
63 }
64 };
65
66 int main ()
67 {
68 Solution testSolution;
69 int count = 5;
70
71 for (int k = 0; k <= count; k ++)
72 {
73 ListNode* pHead = NULL;
74 ListNode* pCur = NULL;
75 ListNode* pDel = NULL;
76
77 for (int i = 0; i < count; i ++)
78 {
79 ListNode* pTemp = new ListNode(i);
80
81 if (i == 0)
82 pHead = pCur = pTemp;
83 else {
84 pCur->next = pTemp;
85 pCur = pCur->next; // pCur->next == pTemp
86 }
87
88 if (i == k)
89 pDel = pTemp;
90 }
91
92 cout << "pHead = " << pHead << endl;
93 if (pHead != NULL)
94 cout << "pHead->val = " << pHead->val << endl;
95 cout << "pDel = " << pDel << endl;
96 if (pDel != NULL)
97 cout << "pDel->val = " << pDel->val << endl;
98 cout << ((pHead == pDel) ? "pHead == pDel" : "pHead != pDel") << endl;
99 cout << endl;
100
101 // Print Linkedlist
102 testSolution.PrintListNode(pHead);
103
104 testSolution.DeleteNode(pHead, pDel);
105 // testSolution.Test(pHead);
106
107 /*
108 delete pDel;
109 pDel = NULL;
110 pHead = NULL;
111 */
112 cout << "pHead = " << pHead << endl;
113 if (pHead != NULL)
114 cout << "pHead->val = " << pHead->val << endl;
115 cout << "pDel = " << pDel << endl;
116 if (pDel != NULL)
117 cout << "pDel->val = " << pDel->val << endl;
118 cout << endl;
119
120 // Print Linkedlist
121 testSolution.PrintListNode(pHead);
122 }
123
124 getchar();
125
126 return 0;
127 }View Code
5:找到链表倒数第K个节点; K=0时，返回链表尾元素。
- 两个指针同时往后走，之间相隔K个元素。一个到达尾部时，另一个正好为倒数第K个。
- 注意边界情况，头指针为NULL，K为0，K大于链表长度。
1 #include <iostream>
2 using namespace std;
3
4 struct ListNode
5 {
6 int val;
7 ListNode *next;
8 ListNode(int x) : val(x), next(NULL) {}
9 };
10
11 class Solution
12 {
13 public:
14 void PrintListNode(ListNode* inListNode)
15 {
16 if (inListNode != NULL) {
17 cout << inListNode->val << endl;
18 PrintListNode(inListNode->next);
19 } else
20 cout << endl;
21 }
22
23 ListNode *findMToLastElement(ListNode* pInHead, int m)
24 {
25 cout << "Entry ListNode *findMToLastElement(ListNode* pInHead, int m)\n" << endl;
26
27 ListNode *pCur = NULL, *pMPre = NULL;
28
29 // Corner case
30 if (pInHead == NULL) return NULL;
31
32 pCur = pMPre = pInHead;
33
34 // Move pCur forward m elements
35 for (int i = 0; i < m; i ++)
36 {
37 if (pCur->next != NULL)
38 pCur = pCur->next;
39 else
40 return NULL;
41 }
42
43 // Move forward together until pCur reach the tail
44 while (pCur->next != NULL)
45 {
46 pCur = pCur->next;
47 pMPre = pMPre->next;
48 }
49
50 // Now pMPre points to the M to last element
51 return pMPre;
52 }
53 };
54
55 int main ()
56 {
57 Solution testSolution;
58 int count = 10;
59
60 ListNode *pHead = NULL;
61 ListNode *pCur = NULL;
62
63 for (int i = 0; i < count; i ++)
64 {
65 ListNode *pTemp = new ListNode(i);
66
67 if (i == 0)
68 pHead = pCur = pTemp;
69 else {
70 pCur->next = pTemp;
71 pCur = pCur->next; // pCur->next == pTemp
72 }
73 }
74
75 // Print Head
76 cout << "pHead = " << pHead << endl;
77 if (pHead != NULL)
78 cout << "pHead->val = " << pHead->val << endl;
79 cout << endl;
80
81 // Print Linkedlist
82 cout << "PrintListNode(pHead)" << endl;
83 testSolution.PrintListNode(pHead);
84
85 cout << "Test findMToLastElement(pHead, 0)\n" << endl;
86 ListNode *pMToLastElement = testSolution.findMToLastElement(pHead, 0);
87
88 // Print Head
89 cout << "pHead = " << pHead << endl;
90 if (pHead != NULL)
91 cout << "pHead->val = " << pHead->val << endl;
92 cout << endl;
93
94 // Print M to last element
95 cout << "pMToLastElement = " << pMToLastElement << endl;
96 if (pMToLastElement != NULL)
97 cout << "pMToLastElement->val = " << pMToLastElement->val << endl;
98 cout << endl;
99
100 getchar();
101
102 return 0;
103 }View Code
6:反转单链表;
- 注意保留反转时，先保留next指针地址，反转之后再赋值给current指针往后走，否则丢了next指针。
1 #include <iostream>
2 using namespace std;
3
4 struct ListNode
5 {
6 int val;
7 ListNode *next;
8 ListNode(int x = 0) : val(x), next(NULL) {}
9 };
10
11 class Solution
12 {
13 public:
14 void PrintList(ListNode *inListNode)
15 {
16 if (inListNode != NULL) {
17 cout << inListNode->val << endl;
18 PrintList(inListNode->next);
19 } else
20 cout << "NULL\n" << endl;
21 }
22
23 // Assign in ptr of the head ptr
24 void DeleteList(ListNode **pInHead)
25 {
26 cout << "Entry void DeleteList(ListNode **pInHead)\n" << endl;
27
28 ListNode *pDel = *pInHead;
29
30 while (pDel != NULL)
31 {
32 // Reserve the address of the next node
33 ListNode *pNext = pDel->next;
34 // Delete the current node
35 delete pDel;
36 pDel = NULL;
37 // Move to the next node
38 pDel = pNext;
39 }
40
41 // Set head to NULL
42 *pInHead = NULL;
43 }
44
45 ListNode* ReverseList(ListNode* pHead)
46 {
47 cout << "Entry ListNode* ReverseList(ListNode* pHead)\n" << endl;
48
49 // Corner case
50 if (pHead == NULL) return NULL;
51
52 ListNode *pPre = NULL, *pRHead = NULL, *pCur = pHead;
53
54 while (pCur != NULL)
55 {
56 // Reserve the next node
57 ListNode *pNext = pCur->next;
58 // Reserve the address of the last node of original list
59 if (pNext == NULL) pRHead = pCur;
60 // Insert current node
61 pCur->next = pPre;
62 // Move the previous ptr
63 pPre = pCur;
64 // Assign the next node back
65 pCur = pNext;
66 }
67
68 return pRHead;
69 }
70 };
71
72 int main ()
73 {
74 Solution testSolution;
75 int count = 10;
76
77 ListNode *pMHead = NULL;
78 ListNode *pMCur = NULL;
79
80 for (int i = 0; i < count; i ++)
81 {
82 ListNode *pTemp = new ListNode(i);
83
84 if (i == 0)
85 pMHead = pMCur = pTemp;
86 else {
87 pMCur->next = pTemp;
88 pMCur = pMCur->next; // pCur->next == pTemp
89 }
90 }
91
92 // Print Head
93 cout << "pMHead = " << pMHead << endl;
94 if (pMHead != NULL)
95 cout << "pMHead->val = " << pMHead->val << endl;
96 cout << endl;
97
98 // Print Linkedlist
99 cout << "PrintList(pMHead)" << endl;
100 testSolution.PrintList(pMHead);
101
102 cout << "Test ReverseList(pMHead)\n" << endl;
103 ListNode *pMRHead = testSolution.ReverseList(pMHead);
104
105 // Print Reversed Head
106 cout << "pMRHead = " << pMRHead << endl;
107 if (pMRHead != NULL)
108 cout << "pMRHead->val = " << pMRHead->val << endl;
109 cout << endl;
110
111 // Print Linkedlist
112 cout << "PrintList(pMRHead)" << endl;
113 testSolution.PrintList(pMRHead);
114
115 // Print Head
116 cout << "pMHead = " << pMHead << endl;
117 if (pMHead != NULL)
118 cout << "pMHead->val = " << pMHead->val << endl;
119 cout << endl;
120
121 // Print Linkedlist
122 cout << "PrintList(pMHead)" << endl;
123 testSolution.PrintList(pMHead);
124
125 cout << "DeleteList(&pMRHead)" << endl;
126 testSolution.DeleteList(&pMRHead);
127
128 // Print Reversed Head
129 cout << "pMRHead = " << pMRHead << endl;
130 if (pMRHead != NULL)
131 cout << "pMRHead->val = " << pMRHead->val << endl;
132 cout << endl;
133
134 // Print Linkedlist
135 cout << "PrintList(pMRHead)" << endl;
136 testSolution.PrintList(pMRHead);
137
138 // Print Head
139 cout << "pMHead = " << pMHead << endl; // It is not NULL, the space is free'ed?
140 if (pMHead != NULL)
141 cout << "pMHead->val = " << pMHead->val << endl;
142 cout << endl;
143
144 // Print Linkedlist
145 // cout << "PrintList(pMHead)" << endl;
146 // testSolution.PrintList(pMHead);
147
148 getchar();
149
150 return 0;
151 }View Code
7:通过两个栈实现一个队列;
- stack 类
- https://msdn.microsoft.com/zh-cn/library/56fa1zk5.aspx
- stack - C++ Reference
- http://www.cplusplus.com/reference/stack/stack/
- Stack_百度百科
- http://baike.baidu.com/link?url=umZEKUINQtIeQve3VZ9u7eJSI2IvARxNaxZus-XwejEcV2QSUjgD-qNQS-efMtjl_96xGF_LXYyUuGO0-ewAzq#5
1 #include <iostream>
2 #include <stack>
3 using namespace std;
4
5 class Solution
6 {
7 public:
8 void push(int node)
9 {
10 stack1.push(node);
11 }
12
13 int pop()
14 {
15 int result = 0;
16
17 if (stack2.empty())
18 {
19 while (!stack1.empty())
20 {
21 stack2.push(stack1.top());
22 stack1.pop();
23 }
24 }
25
26 result = stack2.top();
27 stack2.pop();
28
29 return result;
30 }
31
32 private:
33 stack<int> stack1;
34 stack<int> stack2;
35 };
36
37 int main ()
38 {
39 Solution testSolution;
40 int count = 10;
41
42 // Case 1
43 cout << "CASE 1\n" << endl;
44
45 for (int i = 0; i < count; i ++)
46 {
47 testSolution.push(i);
48 }
49
50 for (int i = 0; i < count; i ++)
51 {
52 cout << testSolution.pop() << endl;
53 }
54
55 // Case 2
56 cout << "\nCASE 2\n" << endl;
57
58 for (int i = 0; i < count; i ++)
59 {
60 testSolution.push(i);
61 cout << testSolution.pop() << endl;
62 }
63
64 // Case 3
65 cout << "\nCASE 3\n" << endl;
66
67 for (int i = 0; i < count / 2; i ++)
68 {
69 testSolution.push(i);
70 }
71
72 cout << testSolution.pop() << "\n" << endl;
73
74 for (int i = count / 2; i < count; i ++)
75 {
76 testSolution.push(i);
77 }
78
79 for (int i = 0; i < count - 1; i ++)
80 {
81 cout << testSolution.pop() << endl;
82 }
83
84 getchar();
85
86 return 0;
87 }View Code
8:二分查找;
- 用vector实现递归和非递归二分查找。刚开始没弄清元素个数，初始化1个元素，再push_back n-1个元素，这时vector里应该是n个元素。
- iterator作为入参传递有问题，先用int代替了，以后再研究
- 用distance来计算找到的当前元素是第几个元素
- vector的二分查找算法 - 轻典 - 博客园
- http://www.cnblogs.com/tianyajuanke/archive/2012/05/23/2515117.html
- distance - C++ Reference
- http://www.cplusplus.com/reference/iterator/distance/
- distance
- https://msdn.microsoft.com/zh-cn/library/k1kfxk7e.aspx
1 #include <iostream>
2 #include <vector>
3 #include <iterator>
4 using namespace std;
5
6 class Solution
7 {
8 public:
9 int binarySearch(vector<int> array, int target)
10 {
11 vector<int>::iterator vecBIT = array.begin();
12 vector<int>::iterator vecEIT = array.end();
13 vector<int>::iterator vecIT = vecBIT + (vecEIT - vecBIT) / 2;
14
15 while (vecBIT <= vecEIT)
16 {
17 if (*vecIT == target)
18 return distance(array.begin(), vecIT); // Return distance between iterators
19 else if (*vecIT < target)
20 {
21 vecBIT = vecIT + 1;
22 vecIT = vecBIT + (vecEIT - vecBIT) / 2;
23 }
24 else
25 {
26 vecEIT = vecIT - 1;
27 vecIT = vecBIT + (vecEIT - vecBIT) / 2;
28 }
29 }
30
31 if (vecBIT > vecEIT) return -1;
32 }
33
34 int binarySearchRecursion(vector<int> array, int vecBIT, int vecEIT, int target)
35 {
36 // cout << "Entry int binarySearchRecursion(vector<int> array, vector<int>::iterator vecBIT, vector<int>::iterator vecEIT, int target)\n" << endl;
37
38 if (vecBIT <= vecEIT)
39 {
40 int vecIT = vecBIT + (vecEIT - vecBIT) / 2;
41
42 if (array[vecIT] == target)
43 return vecIT;
44 else if (array[vecIT] < target)
45 return binarySearchRecursion(array, vecIT + 1, vecEIT, target);
46 else
47 return binarySearchRecursion(array, vecBIT, vecIT - 1, target);
48 }
49 else
50 return -1;
51 }
52 };
53
54 int main ()
55 {
56 Solution testSolution;
57 int n = 3;
58 vector<int> searchArray(1); // initialize vector of n elements with 0
59
60 // push_back another n - 1 elements
61 for (int i = 1; i < n; i ++)
62 searchArray.push_back(i);
63
64 // n element in vector now
65 for (vector<int>::const_iterator vecIT = searchArray.begin(); vecIT < searchArray.end(); vecIT ++)
66 cout << *vecIT << endl;
67 cout << endl;
68
69 cout << distance(searchArray.begin(), searchArray.end()) << "\n" << endl;
70
71 for (int i = 0; i <= n; i ++)
72 {
73 cout << testSolution.binarySearch(searchArray, i) << endl;
74 cout << testSolution.binarySearchRecursion(searchArray, 0, n, i) << endl;
75 }
76
77 getchar();
78
79 return 0;
80 }View Code
9:快速排序;
- 注意对vector入参要用引用，不能用形参，否则对vector元素排序不起作用
- vector有序情况下，快排时间复杂度最高退化为O(N^2)
- 输出vector也可以使用范围for语句
- for (auto vecCIT : vecArray) cout << vecCIT << endl;
1 #include <iostream>
2 #include <vector>
3 using namespace std;
4
5 class Solution
6 {
7 public:
8 // Pass in reference of vector parameter
9 void quickSort(vector<int> &a, int inLeft, int inRight)
10 {
11 cout << "Entry void qSort(vector<int> array, int left, int right)\n" << endl;
12
13 if (inLeft >= inRight)
14 return;
15
16 int i = inLeft, j = inRight, temp = a[inLeft];
17
18 while (i < j)
19 {
20 // Scan from right to left
21 while ((i < j) && (a[j] >= temp))
22 j --;
23 if (i < j)
24 a[i ++] = a[j];
25
26 // Scan from left to right
27 while ((i < j) && (a[i] <= temp))
28 i ++;
29 if (i < j)
30 a[j --] = a[i];
31 }
32 // Base value temp is on the right position
33 a[i] = temp;
34
35 // Divide and conquer
36 quickSort(a, inLeft, i - 1);
37 quickSort(a, i + 1, inRight);
38 }
39 };
40
41 int main ()
42 {
43 Solution testSolution;
44 int array[] = {0, 1, 5, 9, 0, 2, 6, 7, 8, 3, 4};
45 // int array[] = {0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
46 // int array[] = {9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 0};
47 // int array[] = {9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9};
48 vector<int> vecArray(array, array + 11);
49
50 for (vector<int>::const_iterator vecCIT = vecArray.begin(); vecCIT != vecArray.end(); vecCIT ++)
51 cout << *vecCIT << endl;
52 cout << endl;
53
54 testSolution.quickSort(vecArray, 0, vecArray.size() - 1);
55
56 for (vector<int>::const_iterator vecCIT = vecArray.begin(); vecCIT != vecArray.end(); vecCIT ++)
57 cout << *vecCIT << endl;
58 cout << endl;
59
60 getchar();
61
62 return 0;
63 }View Code
10:获得一个int型的数中二进制中1的个数; 输入一个整数，输出该数二进制表示中1的个数。其中负数用补码表示。
- 位运算
- 如果不是负数的话，可以一直右移获取1的个数，f(n) = f(n / 2) + n % 2。跟LeetCode 338 Counting Bits一样。
- http://www.cnblogs.com/pegasus923/p/5499737.html
- 负数用补码表示。所以只能用n = n & (n - 1)，用n跟n-1做&运算，把二进制中的1从后往前去掉，直到没有。注意2的幂的特征(n & (n - 1)) == 0，e.g. 0x1000 & 0x0111。
- 补码_百度百科
- http://baike.baidu.com/link?url=ITDahaDRGLVt2jyxtz0t3a7DbXYFBk1kVZMxzX7SsmxRnTpPBYKHCBYM2zFs1t3s6FrDlgFoYqtTZnnbWClt6a
1 #include <iostream>
2 using namespace std;
3
4 class Solution
5 {
6 public:
7 int NumberOf1(int n)
8 {
9 cout << "Entry int NumberOf1(int n)\n" << endl;
10
11 if (n == 0) return 0;
12
13 int count = 0;
14
15 while (n != 0)
16 {
17 n = n & (n - 1);
18 count ++;
19 }
20
21 return count;
22 }
23 };
24
25 int main ()
26 {
27 Solution testSolution;
28 int n = 10;
29
30 for (int i = 0; i < n; i ++)
31 cout << testSolution.NumberOf1(i) << endl;
32
33 getchar();
34
35 return 0;
36 }View Code
11:输入一个数组，实现一个函数，让所有奇数都在偶数前面; 并保证奇数和奇数，偶数和偶数之间的相对位置不变。
- 跟排序思想差不多。原文的思想类似快排，两个指针一个从头往后找偶数，另一个从尾往前找奇数，找到就交换，直到两个指针相遇。但这个是不稳定的，所以奇偶数之间的相对位置会变。可以用归并，冒泡排序思想，这些都是稳定的。
- 另一种解法就是借助辅助vector存储偶数，扫描一遍把偶数挑出来并从原vector删除，之后再push_back回去。
- 注意vector.end()并不是最后一个元素，而是vector结束符。要找最后一个元素的下标是vector.end()-1。
1 #include <iostream>
2 #include <vector>
3 using namespace std;
4
5 class Solution
6 {
7 public:
8 void reOrderArray(vector<int> &array)
9 {
10 cout << "Entry void reOrderArray(vector<int> &array)\n" << endl;
11
12 vector<int> vecEven; // Store even
13
14 for (vector<int>::iterator vecIT = array.begin(); vecIT != array.end(); )
15 if ((*vecIT % 2) == 0)
16 {
17 vecEven.push_back(*vecIT);
18 vecIT = array.erase(vecIT);
19 }
20 else
21 vecIT ++;
22
23 for (vector<int>::const_iterator vecCIT = vecEven.begin(); vecCIT != vecEven.end(); vecCIT ++)
24 array.push_back(*vecCIT);
25 }
26 };
27
28 int main ()
29 {
30 Solution testSolution;
31 int array[] = {0, 1, 5, 9, 2, 6, 7, 8, 3, 4};
32 vector<int> vecArray(array, array + 10);
33
34 for (vector<int>::const_iterator vecCIT = vecArray.begin(); vecCIT != vecArray.end(); vecCIT ++)
35 cout << *vecCIT << endl;
36 cout << endl;
37
38 testSolution.reOrderArray(vecArray);
39
40 for (vector<int>::const_iterator vecCIT = vecArray.begin(); vecCIT != vecArray.end(); vecCIT ++)
41 cout << *vecCIT << endl;
42 cout << endl;
43
44 getchar();
45
46 return 0;
47 }View Code
12:判断一个字符串是否是另一个字符串的子串;
- 简单的暴力匹配。关键是熟悉string函数，不出BUG。
- kmp算法_百度百科
- http://baike.baidu.com/view/659777.htm
- 字符串匹配的KMP算法_知识库_博客园
- http://kb.cnblogs.com/page/176818/
- 字符串匹配的Boyer-Moore算法_知识库_博客园
- http://kb.cnblogs.com/page/176945/
- KMP算法详解
- http://www.matrix67.com/blog/archives/115
1 #include <iostream>
2 #include <string>
3 using namespace std;
4
5 class Solution {
6 public:
7 int substr(const string source, const string sub)
8 {
9 // Corner case
10 if (sub.empty()) return 0;
11 if (source.empty()) return -1;
12
13 int n = source.length() - sub.length() + 1;
14
15 // better use < rather than <=
16 for (int i = 0; i < n; i ++)
17 {
18 bool flag = true;
19
20 for (int j = 0; j < sub.length(); j ++)
21 {
22 if (source.at(i + j) != sub.at(j))
23 {
24 flag = false;
25 break;
26 }
27 }
28
29 if (flag) return i;
30 }
31
32 return -1;
33 }
34 };
35
36 int main ()
37 {
38 Solution testSolution;
39 string s1 = "mississippi";
40 string s2 = "issi";
41
42 cout << testSolution.substr(s1, s2) << endl;
43
44 getchar();
45
46 return 0;
47 }View Code
13:把一个int型数组中的数字拼成一个串，这个串代表的数字最小;
- 把数组转成字符串然后进行升序排序，定义如果AB < BA，那么A < B。排序完成，再把数组组合就成了最小串了。
- 学习了sort的比较函数的写法。注意类中自定义的compare需要定义成static，否则sort函数编译出错。
- 学习了int跟string转换的几种方法，to_string方便只是是C++11的。
- sort - C++ Reference
- http://www.cplusplus.com/reference/algorithm/sort/?kw=sort
- sort
- https://msdn.microsoft.com/zh-cn/library/ecdecxh1(v=vs.110).aspx
- STL 中sort、qsort 的用法 - rattles的专栏 - 博客频道 - CSDN.NET
- http://blog.csdn.net/rattles/article/details/5510919
- C++中Static作用和使用方法
- http://blog.csdn.net/artechtor/article/details/2312766
- sprintf - C++ Reference
- http://www.cplusplus.com/reference/cstdio/sprintf/
- sprintf、_sprintf_l、swprintf、_swprintf_l、__swprintf_l
- https://msdn.microsoft.com/zh-cn/library/ybk95axf.aspx
- to_string - C++ Reference
- http://www.cplusplus.com/reference/string/to_string/
- C++ int与string的转化 - Andy Niu - 博客园
- http://www.cnblogs.com/nzbbody/p/3504199.html
1 #include <iostream>
2 #include <vector>
3 #include <cstdio> // sprintf
4 #include <algorithm> // sort
5 using namespace std;
6
7 class Solution
8 {
9 public:
10 // Need to define static
11 static bool compare(const string &sA, const string &sB)
12 {
13 string sAB = sA + sB;
14 string sBA = sB + sA;
15
16 return sAB < sBA;
17 }
18
19 string PrintMinNumber(vector<int> numbers)
20 {
21 cout << "Entry string PrintMinNumber(vector<int> numbers)\n" << endl;
22
23 string result = "";
24
25 // Corner case
26 if (numbers.size() == 0) return result;
27
28 vector<string> sNumbers;
29 char *sTemp = new char[10]; // Remember to initialize ptr
30
31 for (vector<int>::const_iterator vecCIT = numbers.begin(); vecCIT != numbers.end(); vecCIT ++)
32 {
33 // string sTemp = to_string(*vecCIT); // C++11
34 sprintf(sTemp, "%d", *vecCIT);
35 sNumbers.push_back(sTemp);
36 }
37
38 // Use self-defined compare function
39 sort(sNumbers.begin(), sNumbers.end(), compare);
40
41 for (vector<string>::const_iterator vecCIT = sNumbers.begin(); vecCIT != sNumbers.end(); vecCIT ++)
42 result += *vecCIT;
43
44 return result;
45 }
46 };
47
48 int main ()
49 {
50 Solution testSolution;
51 int array[] = {3, 32, 321};
52 vector<int> vecArray(array, array + 3);
53 // vector<int> vecArray;
54
55 for (vector<int>::const_iterator vecCIT = vecArray.begin(); vecCIT != vecArray.end(); vecCIT ++)
56 cout << *vecCIT << endl;
57 cout << endl;
58
59 cout << testSolution.PrintMinNumber(vecArray) << endl;
60
61 getchar();
62
63 return 0;
64 }View Code
14:输入一颗二叉树，输出它的镜像（每个节点的左右子节点交换位置）;
- 交换左右子节点（注意是节点，不只是值），然后递归处理左右子树。
- 复习了建树过程。
- swap - C++ Reference
- http://www.cplusplus.com/reference/utility/swap/?kw=swap
1 #include <iostream>
2 #include <vector>
3 using namespace std;
4
5 struct TreeNode {
6 int val;
7 struct TreeNode *left;
8 struct TreeNode *right;
9 TreeNode(int x) :
10 val(x), left(NULL), right(NULL) {
11 }
12 };
13
14 class Solution
15 {
16 public:
17 TreeNode* CreateTree(const vector<int> &vec, const int &pos)
18 {
19 // cout << "Entry TreeNode* CreateTree(const vector<int> &vec, const int &pos)\n" << endl;
20
21 // cout << "pos = " << pos << endl;
22 // cout << "vec.size() = " << vec.size() << endl;
23
24 // Pay attention to >=, or else infinite loop
25 if (pos >= vec.size()) return NULL;
26
27 // cout << "vec[pos] = " << vec[pos] << endl;
28
29 TreeNode *pCur = new TreeNode(vec[pos]);
30 // cout << "pCur->val = " << pCur->val<< endl;
31 pCur->left = CreateTree(vec, 2 * pos);
32 pCur->right = CreateTree(vec, 2 * pos + 1);
33
34 return pCur;
35 }
36
37 void PrintTree(TreeNode *pTreeNode)
38 {
39 if (pTreeNode != NULL) {
40 cout << pTreeNode->val << endl;
41 PrintTree(pTreeNode->left);
42 PrintTree(pTreeNode->right);
43 }
44 }
45
46 void Mirror(TreeNode *pRoot)
47 {
48 // cout << "Entry void Mirror(TreeNode *pRoot) " << endl;
49
50 // Corner case
51 if ((pRoot == NULL) || ((pRoot->left == NULL) && (pRoot->right == NULL))) return;
52
53 // swap two ptr rathet than value
54 swap(pRoot->left, pRoot->right);
55
56 Mirror(pRoot->left);
57 Mirror(pRoot->right);
58 }
59 };
60
61 int main ()
62 {
63 Solution testSolution;
64 int count = 8;
65 int array[] = {0, 8, 6, 10, 5, 7, 9, 11};
66 vector<int> vecArray(array, array + count);
67
68 TreeNode *pMRoot = testSolution.CreateTree(vecArray, 1);
69 /*
70 cout << root << endl;
71 cout << root->val << endl;
72 */
73 testSolution.PrintTree(pMRoot);
74 cout << endl;
75
76 testSolution.Mirror(pMRoot);
77
78 testSolution.PrintTree(pMRoot);
79 cout << endl;
80
81 getchar();
82
83 return 0;
84 }View Code
15:输入两个链表，找到它们第一个公共节点;
- 若有公共节点，则从公共节点之后的所有节点都是公共的。I.e. 短链表是长链表的子串。基于这个思想，长链表里的公共节点只可能出现在倒数K个节点里（K为短链表长度）。长链表先走N-K步，长度相同时再开始往后挨个找。
1 #include <iostream>
2 using namespace std;
3
4 struct ListNode
5 {
6 int val;
7 ListNode *next;
8 ListNode(int x = 0) : val(x), next(NULL) {}
9 };
10
11 class Solution
12 {
13 public:
14 void PrintList(ListNode *pListNode)
15 {
16 if (pListNode != NULL) {
17 cout << pListNode->val << endl;
18 PrintList(pListNode->next);
19 } else
20 cout << "NULL\n" << endl;
21 }
22
23 int getLengthOfList(const ListNode *pListNode)
24 {
25 int length = 0;
26
27 while (pListNode != NULL)
28 {
29 pListNode = pListNode->next;
30 length ++;
31 }
32
33 return length;
34 }
35
36 ListNode* FindFirstCommonNode(ListNode *pHead1, ListNode *pHead2)
37 {
38 // cout << "Entry ListNode* FindFirstCommonNode(ListNode *pHead1, ListNode *pHead2)" << endl;
39
40 int len1 = getLengthOfList(pHead1);
41 int len2 = getLengthOfList(pHead2);
42
43 if (len1 > len2)
44 {
45 int gap = len1 - len2;
46 for (int i = 0; i < gap; i ++)
47 pHead1 = pHead1->next;
48 }
49 else
50 {
51 int gap = len2 - len1;
52 for (int i = 0; i < gap; i ++)
53 pHead2 = pHead2->next;
54 }
55
56 while ((pHead1 != NULL) && (pHead2 != NULL) && (pHead1 != pHead2))
57 {
58 pHead1 = pHead1->next;
59 pHead2 = pHead2->next;
60 }
61
62 return pHead1;
63 }
64 };
65
66 int main ()
67 {
68 Solution testSolution;
69 int count = 10, k = 0;
70
71 ListNode *pMHead = NULL;
72 ListNode *pMCur = NULL;
73 ListNode *pKNode = NULL;
74
75 for (int i = 0; i < count; i ++)
76 {
77 ListNode *pTemp = new ListNode(i);
78
79 if (i == 0)
80 pMHead = pMCur = pTemp;
81 else {
82 pMCur->next = pTemp;
83 pMCur = pMCur->next; // pMCur->next == pTemp
84 }
85
86 if (i == k)
87 pKNode = pMCur;
88 }
89
90 // Print Head
91 cout << "pMHead = " << pMHead << endl;
92 if (pMHead != NULL)
93 cout << "pMHead->val = " << pMHead->val << endl;
94 cout << endl;
95
96 // Print Kth Node
97 cout << "pKNode = " << pKNode << endl;
98 if (pKNode != NULL)
99 cout << "pKNode->val = " << pKNode->val << endl;
100 cout << endl;
101
102 // Print Linkedlist
103 cout << "PrintList(pMHead)" << endl;
104 testSolution.PrintList(pMHead);
105
106 ListNode *pTarget = testSolution.FindFirstCommonNode(pMHead, pKNode);
107
108 // Print Head
109 cout << "pMHead = " << pMHead << endl;
110 if (pMHead != NULL)
111 cout << "pMHead->val = " << pMHead->val << endl;
112 cout << endl;
113
114 // Print Kth Node
115 cout << "pTarget = " << pTarget << endl;
116 if (pTarget != NULL)
117 cout << "pTarget->val = " << pTarget->val << endl;
118 cout << endl;
119
120 // Print Linkedlist
121 cout << "PrintList(pMHead)" << endl;
122 testSolution.PrintList(pMHead);
123
124 getchar();
125
126 return 0;
127 }View Code
转载于:https://www.cnblogs.com/pegasus923/p/5574944.html
学习笔记之15道简单算法题相关推荐
- JAVA 判断简单密码算法_十道简单算法题二【Java实现】
前言 清明不小心就拖了两天没更了-- 这是十道算法题的第二篇了-上一篇回顾:十道简单算法题 最近在回顾以前使用C写过的数据结构和算法的东西,发现自己的算法和数据结构是真的薄弱,现在用Java改写一下, ...
- 十道简单算法题二【Java实现】
前言 清明不小心就拖了两天没更了-- 这是十道算法题的第二篇了-上一篇回顾:十道简单算法题 最近在回顾以前使用C写过的数据结构和算法的东西,发现自己的算法和数据结构是真的薄弱,现在用Java改写一下, ...
- 数据结构学习笔记（六）链表算法题
假期结束,看点题目. 第一题 问题 设顺序表用数组A[]表示,表中元素存储在数组下标1~m+n的范围内,前m个元素递增有序,后n个元素递增有序,设计一个算法,使得整个顺序表有序. (1)给出算法的基本 ...
- java的简单算法题_[2]十道算法题【Java实现】
前言 清明不小心就拖了两天没更了-- 这是十道算法题的第二篇了-上一篇回顾:十道简单算法题 最近在回顾以前使用C写过的数据结构和算法的东西,发现自己的算法和数据结构是真的薄弱,现在用Java改写一下, ...
- 惊！面试现场，简单几道java算法题，90%程序员没写出来
近几个月很多大学实习生离校找实习工作去了,本以为自己很值钱,实际上发现自己并不值钱,想象着自己能赚很多钱了,结果只能拿到一点微薄的工资,很多人会怪公司给的少,但是你有没有想过,你自己的技术值多少钱?你 ...
- tensorflow学习笔记二——建立一个简单的神经网络拟合二次函数
tensorflow学习笔记二--建立一个简单的神经网络 2016-09-23 16:04 2973人阅读 评论(2) 收藏 举报 分类: tensorflow(4) 目录(?)[+] 本笔记目的 ...
- python全排序算法题_Python的100道经典算法题（1）
按照c语言的100道经典算法题,自己原创写的,就得是自己的练习题了 [程序1] 题目:有1.2.3.4个数字,能组成多少个互不相同且无重复数字的三位数?都是多少? 程序分析:可填在百位.十位.个位的数 ...
- python学习笔记（15）循环设计
python学习笔记(15)循环设计 原链:http://www.cnblogs.com/vamei/archive/2012/07/09/2582435.html 注意:zip()在python2 ...
- 机器学习实战（Machine Learning in Action）学习笔记————06.k-均值聚类算法（kMeans）学习笔记...
机器学习实战(Machine Learning in Action)学习笔记----06.k-均值聚类算法(kMeans)学习笔记 关键字:k-均值.kMeans.聚类.非监督学习 作者:米仓山下 时 ...
最新文章
- 台3岁女童疑把玩风枪致死案疑点多 警方将调查厘清
- 上篇文章中ygc越来越慢的case的原因解读
- $.ajax使用总结(一)：Form提交与Payload提交
- 《统计学：从数据到结论》学习笔记(part3)--任何统计量，只要人们觉得合适就可以当成估计量
- 一定要会的synchronized关键字的用法
- java创建和销毁一个对象_有效的Java –创建和销毁对象
- cesium(鼠标事件)
- 北京市委书记蔡奇：加快拓展数字人民币应用全场景试点
- 怎么判断浮点数的有效位数_JavaScript的数据类型及判断
- ARTS打卡计划第一周-Tips-ControllerAdvice的使用
- react实现聊天界面_React-Redux 100行代码简易版探究原理
- jsp 九大内置对象详解以及示例
- VRAR行业深度报告：VRAR是中场，Metaverse是终局
- ROS中使用罗技G29遥控器
- [信息论与编码] 03. 离散信源、信源熵、联合熵、条件熵
- 网上书城项目-LoadRunner压力测试
- C语言学习：原子操作
- js中submit失效
- linux 路由器都会断流吗,有没有迷你路由器推荐？MTK路由器经常断流..
- SCSI的配置全攻略（ISCSI Target/initiator）
热门文章
- spingbot 与 activiti 整个 中创建表而找不到表的问题(创建表失败)
- linux ubuntn j经验
- 前后端分离提交多次验证
- rem适配的浏览器_[史上最全]UI相关尺寸单位详解 | px、pt、dp、sp、rem、vwvh、rpx、ppi、dpi、dppx...
- windows防火墙设置_合理利用Windows 7防火墙，阻止部分功能，避免网络恶意软件攻击...
- 必须声明标量变量 @列名
- oracle xp 安装包资源
- java poi导出Excel表格超大数据量解决方案
- java两个栈怎么完成初始化_Spring bean初始化及销毁你必须要掌握的回调方法
- matlab层次分析法代码_基于主成分分析法和层次分析法的工程项目经理胜任力评价研究...