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文章目录

系列文章目录
一、链表合并
二、士兵队列训练问题
三、Rails
四、Josephus Problem
五、Tree Recovery
六、四则运算
七、愚人节的礼物
八、Web Navigation

一、链表合并

题目描述
输入两个递增排序的链表，合并这两个链表并使新链表中的结点仍然是按照递增排序的。
输入
第一行输入一个数字T表示样例个数。
对于每个样例，第一行输入两个数字m, n，分别表示两个链表的长度
第二行输入m个数字，表示第一个链表
第三行输入n个数字，表示第二个链表
输出
对于每一组样例，输出 m + n个数表示合并结果。
样例输入
2
2 2
1 2
3 4
3 2
3 4 5
1 4
样例输出
1 2 3 4
1 3 4 4 5
提示
m + n <= 1000
题目分析
本题利用数组来实现链表，因为使用链表过于复杂，所用两个数组代替两个链表，并且使用第三个数组储存前两个数组中的元素，最后使用sort函数把第三个数组中的元素从小到大排序，最后输出即可（注意外层循环）

#include <bits/stdc++.h>
using namespace std;
int a[1001], b[1001], c[1001];
int main()
{int T;cin >> T;while (T--){int m, n;cin >> m >> n;for (int i = 0; i < m; i++)
{cin >> a[i];c[i] = a[i];//输入第一个数组中的元素并且存在第三个数组里}for (int j = 0; j < n; j++)
{cin >> b[j];c[m + j] = b[j];//输入第二个数组的元素并且存在第三个数组里}sort(c, c + m + n);//把第三个数组中的元素从小到大排序for (int i = 0; i < m + n; i++)cout << c[i] << " ";//输出cout << endl;}return 0;
}

二、士兵队列训练问题

题目描述
某部队进行新兵队列训练，将新兵从一开始按顺序依次编号，并排成一行横队，训练的规则如下：从头开始一至二报数，凡报到二的出列，剩下的向小序号方向靠拢，再从头开始进行一至三报数，凡报到三的出列，剩下的向小序号方向靠拢，继续从头开始进行一至二报数。。。，以后从头开始轮流进行一至二报数、一至三报数直到剩下的人数不超过三人为止。
输入
本题有多个测试数据组，第一行为组数N，接着为N行新兵人数，新兵人数不超过5000。

输出
共有N行，分别对应输入的新兵人数，每行输出剩下的新兵最初的编号，编号之间有一个空格。

样例输入
2
20
40
样例输出
1 7 19
1 19 37
题目分析
本题的思路就是建立两个队列，每次判断是奇数次还是偶数次转换报数方式，如果是奇数次，那么就进行一二报数，也就是报到二的还在这个队里，所有剩下的元素进入另外一个队里；之后进行执行队列交换，也就是要对储存剩下元素的队列进行一三报数，报到3的留在这个队里，其余所有元素进入另外一个队里。以此类推直到最后剩下不超过三个元素。

#include <bits/stdc++.h>
#define N 5002
using namespace std;
struct Q {int f, r, da[N];void init() {f = r = 0;//初始化队列的头和尾都是零}void push(int a) {da[r++] = a;//把元素a加在队列尾部}int gettop() {return da[f++];//得到队列的第一个元素}bool ch() {return r - f < 4;//判断是否剩下不超过三个人}bool empty() {return f == r;//判断队列是否为空}
}q[2];
int main()
{int n, num, i;cin >> n;while (n--){cin >> num;q[0].init(), q[1].init();//初始化前两个元素for (int i = 1; i <= num; i++)q[0].push(i);//每一个士兵都入队int now = 0, pre = 1, i = 0;        while (!q[now].ch())//当现在剩下的人数比三个人多时{q[pre].init();//初始化队列q[1]if (i & 1) {//当i等于1的时候（也就是该进行一三报数的时候）while (!q[now].empty()) {//当队列q[now]不空时q[pre].push(q[now].gettop());//在队列q[pre]中加入q[now]中的第一个元素if (q[now].empty()) break;//如果q[now]空了就退出循环（仅有一个元素）q[pre].push(q[now].gettop());//不空就继续在q[pre]中加入q[now]第二个元素if (q[now].empty()) break;//如果q[now]空了就退出循环（仅有两个元素）q[now].gettop();//使得q[now]中剩下的元素是报数时报3的元素}//现在得到的q[now]是所有出队的元素，q[pre]里的元素是接下来要继续进行报数的元素}else {//当i等于0的时候，也就是该进行一二报数的时候while (!q[now].empty()) {//当队列q[now]不空的时候q[pre].push(q[now].gettop());//在队列q[pre]中加入q[now]中的第一个元素if (q[now].empty()) break;//如果q[now]空了就退出循环（仅有一个元素）q[now].gettop();//使得q[now]中剩下的元素是报数时报2的元素}//现在得到的q[now]是所有出队的元素，q[pre]里的元素是接下来要继续进行报数的元素}i ^= 1;//改变i的值（如果之前是0就改成1，之前是1就改成0）swap(now, pre);//改变两个队列的索引，也就是如果之前是对q[0]进行取队首操作，}                   //下一次循环就要对q[1]进行取队首操作，交替进行cout << q[now].gettop() << " ";//输出q[0]的队首元素while (!q[now].empty()) {//当队列q[0]不空的时候cout << q[now].gettop() << " ";//输出队列q[0]的队首元素（向后取）}cout << endl;
}
return 0;
}

三、Rails

题目描述
There is a famous railway station in PopPush City. Country there is incredibly hilly. The station was built in last century. Unfortunately, funds were extremely limited that time. It was possible to establish only a surface track. Moreover, it turned out that the station could be only a dead-end one (see picture) and due to lack of available space it could have only one track.
The local tradition is that every train arriving from the direction A continues in the direction B with coaches reorganized in some way. Assume that the train arriving from the direction A has N <= 1000 coaches numbered in increasing order 1, 2, …, N. The chief for train reorganizations must know whether it is possible to marshal coaches continuing in the direction B so that their order will be a1, a2, …, aN. Help him and write a program that decides whether it is possible to get the required order of coaches. You can assume that single coaches can be disconnected from the train before they enter the station and that they can move themselves until they are on the track in the direction B. You can also suppose that at any time there can be located as many coaches as necessary in the station. But once a coach has entered the station it cannot return to the track in the direction A and also once it has left the station in the direction B it cannot return back to the station.

输入
The input consists of blocks of lines. Each block except the last describes one train and possibly more requirements for its reorganization. In the first line of the block there is the integer N described above. In each of the next lines of the block there is a permutation of 1, 2, …, N. The last line of the block contains just 0.

The last block consists of just one line containing 0.
输出
The output contains the lines corresponding to the lines with permutations in the input. A line of the output contains Yes if it is possible to marshal the coaches in the order required on the corresponding line of the input. Otherwise it contains No. In addition, there is one empty line after the lines corresponding to one block of the input. There is no line in the output corresponding to the last ``null’’ block of the input.

样例输入
5
1 2 3 4 5
5 4 1 2 3
0
6
6 5 4 3 2 1
0
0
样例输出
Yes
No

Yes
题目分析
本题的思路就在于栈和数组，在数组中储存火车驶出火车站的顺序，为了模拟火车进站过程，我们需要在for循环下让第i辆火车进栈，同时在while循环下判断进入火车站的火车的序号是否等于栈顶元素，如果相等就驶出，同时驶出火车数加1，数组的索引向后移一位，代表该驶出下一辆火车，此时如果满足while循环的条件就可以一直驶出。（注意在输入有几辆火车以及火车驶出站的次序时判断不是0）

#include <bits/stdc++.h>
using namespace std;
int main()
{stack<int > sta;int n, x[1001], y;while(cin >> n && n)//确保至少通过一辆火车{memset(x, 0, sizeof(x));while(cin >> x[1] && x[1]) {//保证输入的数不是0int k = 0, p = 0, q = 1;for(int i = 2;i <= n; i++)//输入想要驶出站的顺序cin >> x[i];for(int i = 1; i <= n; i++) {sta.push(i);//把进入火车站的顺序存在栈里while(x[q] == sta.top()) {//当栈顶元素等于将要驶出的火车的序号k++, q++;//输出的火车数k加1，将要驶出的火车的索引向后移一位sta.pop();//将要驶出的火车驶出if(sta.empty())   break;//当火车站中没有火车时就退出循环}}if(k == n)    cout << "Yes" << endl;//如果驶出的火车数等于驶入的火车数就输出yeselse    cout << "No" << endl;//否则输出no}cout << endl;//输出空行}return 0;
}

四、Josephus Problem

题目描述
The historian Flavius Josephus relates how, in the Romano-Jewish conflict of 67 A.D., the Romans took the town of Jotapata which he was commanding. Escaping, Josephus found himself trapped in a cave with 40 companions. The Romans discovered his whereabouts and invited him to surrender, but his companions refused to allow him to do so. He therefore suggested that they kill each other, one by one, the order to be decided by lot. Tradition has it that the means for affecting the lot was to stand in a circle, and, beginning at some point, count round, every third person being killed in turn. The sole survivor of this process was Josephus, who then surrendered to the Romans. Which begs the question: had Josephus previously practiced quietly with 41 stones in a dark corner, or had he calculated mathematically that he should adopt the 31st position in order to survive?
Now you are in a similar situation. There are n persons standing in a circle. The persons are numbered from 1 to ncircularly. For example, 1 and n are adjacent and 1 and 2 are also. The count starts from the first person. Each time you count up to k and the kth person is killed and removed from the circle. Then the count starts from the next person. Finally one person remains. Given n and k you have to find the position of the last person who remains alive.
输入
Input starts with an integer T (≤ 200), denoting the number of test cases.
Each case contains two positive integers n (1 ≤ n ≤ 200) and k (1 ≤ k < 201).
输出
For each case, print the case number and the position of the last remaining person.
样例输入
6
2 1
2 2
3 1
3 2
3 3
4 6
样例输出
Case 1: 2
Case 2: 1
Case 3: 3
Case 4: 3
Case 5: 2
Case 6: 3
题目分析
本题属于约瑟夫环，需要理解约瑟夫环的实质，对于一个约瑟夫环，如果输入的数据是10 3，就代表一共有10个人，报数报到三就要死，之后456也报123，相当于都向前移动了三位（当有一个人死了的时候），也就是说，当有9个人的时候，幸存者的序号就等于有十个人的时候幸存者的序号减三；有11个人的时候，幸存者的序号就等于有10个人的时候幸存者的序号加三，于是有了代码中的递归。

#include <bits/stdc++.h>
using namespace std;
int T, n, k;
int main()
{cin >> T;While (T--){cin >> n >> k;//输入人数和报到k时第k个人被杀int res = 0;for(int j = 2; j <= n; j++)res = (res + k) % j;//递归思想printf("Case %d: %d\n",i, res + 1);}return 0;
}

五、Tree Recovery

题目描述
Little Valentine liked playing with binary trees very much. Her favorite game was constructing randomly looking binary trees with capital letters in the nodes.
This is an example of one of her creations:

                                           D/ \/   \B     E/ \     \/   \     \A     C     G//F

To record her trees for future generations, she wrote down two strings for each tree: a preorder traversal (root, left subtree, right subtree) and an inorder traversal (left subtree, root, right subtree). For the tree drawn above the preorder traversal is DBACEGF and the inorder traversal is ABCDEFG.
She thought that such a pair of strings would give enough information to reconstruct the tree later (but she never tried it).
Now, years later, looking again at the strings, she realized that reconstructing the trees was indeed possible, but only because she never had used the same letter twice in the same tree.
However, doing the reconstruction by hand, soon turned out to be tedious.
So now she asks you to write a program that does the job for her!

输入
The input will contain one or more test cases.
Each test case consists of one line containing two strings preord and inord, representing the preorder traversal and inorder traversal of a binary tree. Both strings consist of unique capital letters. (Thus they are not longer than 26 characters.)
Input is terminated by end of file.

输出
For each test case, recover Valentine’s binary tree and print one line containing the tree’s postorder traversal (left subtree, right subtree, root).

样例输入
DBACEGF ABCDEFG
BCAD CBAD
样例输出
ACBFGED
CDAB
题目分析
给定前序和中序遍历结果，输出后序遍历结果。使用深度优先搜索算法，对于前序遍历和中序遍历进行搜索，首先找到根节点，通过递归分别把左子树和右子树的根节点和左子树右子树找出来，最后输出后序遍历结果即可。

#include <bits/stdc++.h>
using namespace std;
int len;
string pre, in;
void dfs(int b1, int e1, int b2, int e2) {//分别代表前序中序遍历的开始和结束位置if (b1 > e1) return;//如果前序遍历的开始大于结束就返回（已经到达树底部）int k = b2;//k = 中序遍历的开始位置while (in[k] != pre[b1]) k++;//如果中序遍历k位置的元素不等于前序遍历的根节点，k++int res = k - b2;//res代表根节点的位置dfs(b1 + 1, b1 + res, b2, b2 + res - 1);//前序遍历的左子树和中序遍历的左子树遍历dfs(b1 + res + 1, e1, k + 1, e2);//前序遍历的右子树和中序遍历的右子树printf("%c", pre[b1]);//在搜索完成之后输出最底层的左子树右子树和根节点
}
int main()
{while(cin >> pre) {cin >> in;len = pre.size();dfs(0, len - 1, 0, len - 1);//对整个前序遍历和中序遍历进行深度优先搜索cout << endl;//输出空行}return 0;
}

六、四则运算

题目描述
输入一行字符串，是包含±/合法运算的一行字符串，要求解析字符串的值。
输入
第一行包括一行整数T，表示样例个数。
对于每一个样例，输入一行字符串代表合法运算。
输出
对于每一个样例，输出一个数字表示结果。结果误差要求小于1e-4。
样例输入
2
3/4+2
2+34
样例输出
2.75
14
题目分析
本题的思路就是建立两个栈，一个是符号栈，另一个是数字栈，在输入字符串之后，判断是数字还是符号，分别进栈。同时如果符号是±，直接进入符号栈，如果是/，则需要弹出数字栈的栈顶元素，并且和字符串中的下一个数字进行计算，之后把结果弹在数字栈中。最后符号栈只剩下±，所以最后需要在while循环里判断当两个栈都不空时，进行加减运算，最后当符号栈空后，弹出数字站栈顶的元素，也就是计算的结果。*

#include <iostream>
#include <stdlib.h>
#include <stdio.h>
using namespace std;
typedef char ElemType;
typedef double ElemType2;
typedef bool Status;
typedef struct SNode
{ElemType data;struct SNode *next;
}SNode, *SLNode;
typedef struct
{SLNode top;int count;
}SLinkStack;
Status Init_LStack(SLinkStack *s)//初始化栈
{s -> top = (SLNode)malloc(sizeof(SNode));//建立栈顶元素的空间if(!s)   return 0;s -> top = NULL;//栈顶元素指向的是空s -> count = 0;//栈内元素的个数是0return 1;
}
Status Push(SLinkStack *s, ElemType e)//进栈作用
{SLNode p;p = (SLNode)malloc(sizeof(SNode));if(!p)   return 0;p -> data = e;p -> next = s -> top;s -> top = p;s -> count++;return 1;
}
Status Pop(SLinkStack *s, ElemType *e)
{SLNode p;if (s -> top == NULL)   return 0;//如果栈空，代表无法弹栈*e = s -> top -> data;//*e代表栈顶元素的值p = s -> top;//p代表栈顶元素s -> top = p -> next;//栈顶元素更新为p的下一个free(p);//释放p的空间s -> count--;//栈内元素减1return 1;
}
typedef struct SNode2
{ElemType2 data;struct SNode2 *next;
}SNode2, *SLNode2;
typedef struct
{SLNode2 top;int count;
}SLinkStack2;
Status Init_LStack2(SLinkStack2 *s)//也是初始化的作用
{s -> top = (SLNode2)malloc(sizeof(SNode2));if (!s)  return 0;s -> top = NULL;s -> count = 0;return 1;
}
Status Push2(SLinkStack2 *s,ElemType2 e)
{SLNode2 p;p = (SLNode2)malloc(sizeof(SNode2));//新建一个结点的空间if (!p)   return 0;p -> data = e;//p的值是ep -> next = s -> top;//p的下一个元素是原来的栈顶元素s -> top = p;//栈顶元素更新为ps -> count++;//栈内元素加1return 1;
}
Status Pop2(SLinkStack2 *s,ElemType2 *e)
{SLNode2 p;if(s -> top == NULL)   return 0;//如果栈顶没有元素则弹出失败*e = s -> top -> data;//*e就代表栈内元素的个数p = s -> top;//p就代表栈顶元素s -> top = p -> next;//栈顶元素后移free(p);//把p所占的空间释放（也就相当于弹栈）s -> count--;//栈顶元素减少1return 1;
}
Status GetTop_LStack2(SLinkStack2 s, ElemType2 *e)
{if(s.top == NULL)  return 0;*e = s.top -> data;return 1;
}
int main()
{int T;string s;double e,e2;char m;cin >> T;SLinkStack2 S1; SLinkStack S2;//建立两个栈（数字和符号）while(T--){Init_LStack2(&S1);Init_LStack(&S2);//初始化两个栈cin >> s;//输入字符串for(unsigned int i = 0; i < s.length(); i++){if(s[i] >= '0'&& s[i] <= '9') {//如果这个字符是数字Push2(&S1, s[i] - '0');//变成数字进栈1}else {//如果是字符if(s[i] == '+' || s[i] == '-')//当是+-时Push(&S2, s[i]);//进入栈2else {//当是*/时Pop2(&S1, &e);//把栈1中的栈顶元素弹出并且计算if(s[i] == '*')e *= (s[i + 1] - '0');elsee /= (s[i + 1] - '0');i++;//继续后一位的输入Push2(&S1, e);//把计算出来的数字弹入栈1}}}while(S1.top != NULL && S2.top != NULL) {//当两个栈都不空时Pop2(&S1, &e2), Pop(&S2, &m), Pop2(&S1, &e);//s1弹栈，s2弹栈，s1弹栈if(m == '+')   e += e2;//如果s2弹出的符号是+，计算else    e-=e2;//否则一定是减Push2(&S1, e);//把计算好的结果弹进栈s1内}GetTop_LStack2(S1, &e);//得到s1的栈顶元素（也就是最终计算结果）cout << e << endl;//输出即可}
}

七、愚人节的礼物

题目描述
四月一日快到了，Vayko想了个愚人的好办法——送礼物。嘿嘿，不要想的太好，这礼物可没那么简单，Vayko为了愚人，准备了一堆盒子，其中有一个盒子里面装了礼物。盒子里面可以再放零个或者多个盒子。假设放礼物的盒子里不再放其他盒子。
用()表示一个盒子，B表示礼物，Vayko想让你帮她算出愚人指数，即最少需要拆多少个盒子才能拿到礼物。

输入
本题目包含多组测试，请处理到文件结束。
每组测试包含一个长度不大于1000,只包含’(’,’)'和’B’三种字符的字符串，代表Vayko设计的礼物透视图。
你可以假设，每个透视图画的都是合法的。
输出
对于每组测试，请在一行里面输出愚人指数。

样例输入
((((B)()))())
(B)
样例输出
4
题目分析
本题的思路就是，在for循环下进行字符串的遍历操作，当遍历到左括号时，直接拆掉的礼物数++，当遍历到右括号时，代表遍历到了一个空盒子，所以count–，代表可以不用拆这个盒子，当遍历到B时，也就是礼物时，就代表已经拆了最少的盒子，可以输出count了。

#include <bits/stdc++.h>
using namespace std;int main()
{string s;while (getline(cin, s))//输入字符串{int count = 0;//count代表最少要拆掉的盒子的个数for (int j = 0; ; j++){if (s[j] == 'B')//当遍历的元素正好是礼物时break;//就代表可以通过至少拆除count个左括号来拿到礼物if (s[j] == '(')//当遍历的元素是左括号时{count ++;//拆掉盒子的个数++}if (s[j] == ')')//当是右括号时，拆掉盒子的个数--{count --;}}cout << count << endl;}return 0;
}

八、Web Navigation

题目描述
Standard web browsers contain features to move backward and forward among the pages recently visited. One way to implement these features is to use two stacks to keep track of the pages that can be reached by moving backward and forward. In this problem, you are asked to implement this. The following commands need to be supported:
BACK: Push the current page on the top of the forward stack. Pop the page from the top of the backward stack, making it the new current page. If the backward stack is empty, the command is ignored.
FORWARD: Push the current page on the top of the backward stack. Pop the page from the top of the forward stack, making it the new current page. If the forward stack is empty, the command is ignored.
VISIT : Push the current page on the top of the backward stack, and make the URL specified the new current page. The forward stack is emptied.
QUIT: Quit the browser.
Assume that the browser initially loads the web page at the URL http://www.acm.org/
输入
Input is a sequence of commands. The command keywords BACK, FORWARD, VISIT, and QUIT are all in uppercase. URLs have no whitespace and have at most 70 characters. You may assume that no problem instance requires more than 100 elements in each stack at any time. The end of input is indicated by the QUIT command.
输出
For each command other than QUIT, print the URL of the current page after the command is executed if the command is not ignored. Otherwise, print “Ignored”. The output for each command should be printed on its own line. No output is produced for the QUIT command.
样例输入
VISIT http://acm.ashland.edu/
VISIT http://acm.baylor.edu/acmicpc/
BACK
BACK
BACK
FORWARD
VISIT http://www.ibm.com/
BACK
BACK
FORWARD
FORWARD
FORWARD
QUIT
样例输出
http://acm.ashland.edu/
http://acm.baylor.edu/acmicpc/
http://acm.ashland.edu/
http://www.acm.org/
Ignored
http://acm.ashland.edu/
http://www.ibm.com/
http://acm.ashland.edu/
http://www.acm.org/
http://acm.ashland.edu/
http://www.ibm.com/
Ignored
题目分析
建立两个栈存放之前的网站和之后的网站，然后设定一个数t代表现在的网站，之后跟随不同的指令，进行弹栈入栈输出更新t的操作即可。

#include <bits/stdc++.h>
#include <stack>
using namespace std;
int main() {string t = "http://www.acm.org/", s, ss;stack <string> ff, bb;//分别代表现在的网站t之后的网站和之前的网站while (!ff.empty()) ff.pop();//初始化，弹空栈while (!bb.empty()) bb.pop();//初始化，弹空栈string v="VISIT", b="BACK", f="FORWARD", q="QUIT", i="Ignored";//简写ff.push(t);//t代表的网站进栈ffwhile (cin >> s) {//输入指令if (s == q)   break;//如果指令是quit，代表退出，直接breakif (s == v) {//如果指令是visitcin >> ss;bb.push(t);//t代表的网站进栈bbwhile(!ff.empty())ff.pop();//弹空ff栈cout << ss << endl;//输出现在的网站sst = ss;//现在的网站t就是ss}if (s == b) {//如果指令是backif (bb.empty())     cout << i << endl;//如果bb栈空，就代表没有之前的网站，输出ielse {//如果bb栈不空ff.push(t);//t进入ff栈（代表前面的网站）t = bb.top();//更新t为bb栈的栈顶网站cout << t << endl;//输出现在的网站tbb.pop();//弹出bb栈的栈顶元素}}if (s == f) {//当指令是forward时if (ff.empty())    cout << i << endl;//当ff栈空，输出ielse {//当ff栈不空时bb.push(t);//把现在的网站t弹入bb栈（已经变成了之前的网站）t = ff.top();//现在的网站t更新为ff栈的栈顶网站cout << t << endl;//输出t也就是现在的网站ff.pop();//弹出ff栈的栈顶元素（因为不是之后的网站，而是现在的网站）}}}return 0;
}

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