【链接】：Two Sum
【描述】：Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
Example: Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. Subscribe to see which companies asked this question. 【思路】： C代码：
int main(){ int n,arr[maxn]; int target; int ri,rj; memset(arr,0,sizeof(arr)); scanf("%d",&n); for(int i=0; i<n; ++i) scanf("%d",&arr[i]); scanf("%d",&target); for(int i=0; i<n; ++i){ for(int j=i+1; j<n; ++j){ if(arr[j]==(target-arr[i])) ri=i; rj=j; break; } break; } PRintf("%d %d/n",ri,rj); return 0;}C++代码：
class Solution{public: vector <int> twoSum(vector<int>&nums,int target) { unordered_map<int,int> map; for(int i=0; i<nums.size(); ++i){ auto tp=map.find(target-nums[i]); if(tp!=map.end()) { return {tp->second,i}; } map[nums[i]]=i; } }};unordered_map和map类似，都是存储的key-value的值，可以通过key快速索引到value。不同的是unordered_map不会根据key的大小进行排序，
存储时是根据key的hash值判断元素是否相同，即unordered_map内部元素是无序的，而map中的元素是按照二叉搜索树存储，进行中序遍历会得到有序遍历。
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