leetcode Two Sum

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.

Example:

Given nums = [2, 7, 11, 15], target = 9,Because nums[0] + nums[1] = 2 + 7 = 9,return [0, 1].方法一：暴力加上稍微的剪枝通过，O(n^2)无法通过class Solution {public:    vector<int> twoSum(vector<int>& nums, int target) {        unordered_map<int, int> m;         vector<int> res(2);        for(int i=0; i<nums.size(); i++)        {            if(m.find(target-nums[i])==m.end())            m[nums[i]]=i;            else            {                res[0]=m[target-nums[i]];                res[1]=i;                break;                           }        }         return res;    }};方法二：一次遍历，使用unordered_map,使用hash索引，此map如下
template < class Key,                                    // unordered_map::key_type           class T,                                      // unordered_map::mapped_type           class Hash = hash<Key>,                       // unordered_map::hasher           class Pred = equal_to<Key>,                   // unordered_map::key_equal           class Alloc = allocator< pair<const Key,T> >  // unordered_map::allocator_type           > class unordered_map;


代码：
class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int, int> m; 
        vector<int> res(2);
        for(int i=0; i<nums.size(); i++)
        {
            if(m.find(target-nums[i])==m.end())
            m[nums[i]]=i;
            else
            {
                res[0]=m[target-nums[i]];
                res[1]=i;
                break;
               
            }
        }
         return res;
    }
};