前言

经典面试题，其实比较简单，但鉴于刷题一遍容易忘，还是记录一下。

一、题目

二、方法

1.暴力破解

暴力破解可以解决很多问题，但是一般来说不太可能通过，因为耗时太长，但怎么说也是一种方法。
代码如下：

int* twoSum(int* nums, int numsSize, int target, int* returnSize){int* ret = NULL;for(int i=0; i<numsSize-1; i++){for(int j=i+1; j<numsSize; j++){if((nums[i]+nums[j])==target){ret = malloc(sizeof(int) * 2);ret[0]=i;ret[1]=j;*returnSize = 2;return (ret);}}}*returnSize = 0;return (returnSize);
}

2.哈希表

C语言实现代码如下：

struct hashTable {int key;int val;UT_hash_handle hh;
};struct hashTable* hashtable;struct hashTable* find(int ikey) {struct hashTable* tmp;HASH_FIND_INT(hashtable, &ikey, tmp);return tmp;
}void insert(int ikey, int ival) {struct hashTable* it = find(ikey);if (it == NULL) {struct hashTable* tmp = malloc(sizeof(struct hashTable));tmp->key = ikey, tmp->val = ival;HASH_ADD_INT(hashtable, key, tmp);} else {it->val = ival;}
}int* twoSum(int* nums, int numsSize, int target, int* returnSize) {hashtable = NULL;for (int i = 0; i < numsSize; i++) {struct hashTable* it = find(target - nums[i]);if (it != NULL) {int* ret = malloc(sizeof(int) * 2);ret[0] = it->val, ret[1] = i;*returnSize = 2;return ret;}insert(nums[i], i);}*returnSize = 0;return NULL;
}

C++语言代码实现如下：

class Solution {public:vector<int> twoSum(vector<int>& nums, int target) {unordered_map<int, int> hashtable;for (int i = 0; i < nums.size(); ++i) {auto it = hashtable.find(target - nums[i]);if (it != hashtable.end()) {return {it->second, i};}hashtable[nums[i]] = i;}return {};}
};

其他一些总结

（1）关于unordered_map

unordered_map记录元素的hash值，根据hash值判断元素是否相同。

效率：无论从查找、插入上来说，unordered_map的效率都优于hash_map，更优于map；
空间复杂度：hash_map最低，unordered_map次之，map最大。

unordered_map感觉速度和hash_map差不多，但是支持string做key，也可以使用复杂的对象作为key。

成员函数：

应用示例1：

 1 // unordered_map::insert2 #include <iostream>3 #include <string>4 #include <unordered_map>5 using namespace std;6 7 void display(unordered_map<string,double> myrecipe,string str)8 {9     cout << str << endl;
10     for (auto& x: myrecipe)
11         cout << x.first << ": " << x.second << endl;
12     cout << endl;
13 }
14
15 int main ()
16 {17     unordered_map<string,double>
18     myrecipe,
19     mypantry = {{"milk",2.0},{"flour",1.5}};
20
21     /****************插入*****************/
22     pair<string,double> myshopping ("baking powder",0.3);
23     myrecipe.insert (myshopping);                        // 复制插入
24     myrecipe.insert (make_pair<string,double>("eggs",6.0)); // 移动插入
25     myrecipe.insert (mypantry.begin(), mypantry.end());  // 范围插入
26     myrecipe.insert ({{"sugar",0.8},{"salt",0.1}});    // 初始化数组插入(可以用二维一次插入多个元素，也可以用一维插入一个元素)
27     myrecipe["coffee"] = 10.0;  //数组形式插入
28
29     display(myrecipe,"myrecipe contains:");
30
31     /****************查找*****************/
32     unordered_map<string,double>::const_iterator got = myrecipe.find ("coffee");
33
34     if ( got == myrecipe.end() )
35         cout << "not found";
36     else
37         cout << "found "<<got->first << " is " << got->second<<"\n\n";
38     /****************修改*****************/
39     myrecipe.at("coffee") = 9.0;
40     myrecipe["milk"] = 3.0;
41     display(myrecipe,"After modify myrecipe contains:");
42
43
44     /****************擦除*****************/
45     myrecipe.erase(myrecipe.begin());  //通过位置
46     myrecipe.erase("milk");    //通过key
47     display(myrecipe,"After erase myrecipe contains:");
48
49     /****************交换*****************/
50     myrecipe.swap(mypantry);
51     display(myrecipe,"After swap with mypantry, myrecipe contains:");
52
53     /****************清空*****************/
54     myrecipe.clear();
55     display(myrecipe,"After clear, myrecipe contains:");
56     return 0;
57 }
58 /*
59 myrecipe contains:
60 salt: 0.1
61 milk: 2
62 flour: 1.5
63 coffee: 10
64 eggs: 6
65 sugar: 0.8
66 baking powder: 0.3
67
68 found coffee is 10
69
70 After modify myrecipe contains:
71 salt: 0.1
72 milk: 3
73 flour: 1.5
74 coffee: 9
75 eggs: 6
76 sugar: 0.8
77 baking powder: 0.3
78
79 After erase myrecipe contains:
80 flour: 1.5
81 coffee: 9
82 eggs: 6
83 sugar: 0.8
84 baking powder: 0.3
85
86 After swap with mypantry, myrecipe contains:
87 flour: 1.5
88 milk: 2
89
90 After clear, myrecipe contains:
91 */

应用示例2：（遍历）

 1 // unordered_map::bucket_count2 #include <iostream>3 #include <string>4 #include <unordered_map>5 using namespace std;6 7 int main ()8 {9     unordered_map<string,string> mymap =
10     {11         {"house","maison"},
12         {"apple","pomme"},
13         {"tree","arbre"},
14         {"book","livre"},
15         {"door","porte"},
16         {"grapefruit","pamplemousse"}
17     };
18     /************begin和end迭代器***************/
19     cout << "mymap contains:";
20     for ( auto it = mymap.begin(); it != mymap.end(); ++it )
21         cout << " " << it->first << ":" << it->second;
22     cout << endl;
23     /************bucket操作***************/
24      unsigned n = mymap.bucket_count();
25
26     cout << "mymap has " << n << " buckets.\n";
27
28     for (unsigned i=0; i<n; ++i)
29     {30         cout << "bucket #" << i << "'s size:"<<mymap.bucket_size(i)<<" contains: ";
31         for (auto it = mymap.begin(i); it!=mymap.end(i); ++it)
32             cout << "[" << it->first << ":" << it->second << "] ";
33         cout << "\n";
34     }
35
36     cout <<"\nkey:'apple' is in bucket #" << mymap.bucket("apple") <<endl;
37     cout <<"\nkey:'computer' is in bucket #" << mymap.bucket("computer") <<endl;
38
39     return 0;
40 }
41 /*
42 mymap contains: door:porte grapefruit:pamplemousse tree:arbre apple:pomme book:livre house:maison
43 mymap has 7 buckets.
44 bucket #0's size:2 contains: [book:livre] [house:maison]
45 bucket #1's size:0 contains:
46 bucket #2's size:0 contains:
47 bucket #3's size:2 contains: [grapefruit:pamplemousse] [tree:arbre]
48 bucket #4's size:0 contains:
49 bucket #5's size:1 contains: [apple:pomme]
50 bucket #6's size:1 contains: [door:porte]
51
52 key:'apple' is in bucket #5
53
54 key:'computer' is in bucket #6
55 */

（以上内容来自：详细介绍C++STL：unordered_map）

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前言