LeetCode–1.TwoSum

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].

首先想到的直接两个循环不就完了，代码如下：

#include<iostream>#include<vector>using namespace std;class Solution{public:    vector<int> twoSum(vector<int>& nums, int target)    {        vector<int> res;        for (int i = 0; i < nums.size() - 1; ++i)        {            for (int j = i + 1; j < nums.size(); ++j)                if (nums[i] + nums[j] == target)                {                    res.push_back(i);                    res.push_back(j);                    break;                }        }        return res;    }};int main(){    vector<int> nums{ 2,7,11,15 };    int target = 9;    Solution s;    vector<int> res = s.twoSum(nums, target);    system("pause");    return 0;}

结果提交了之后超时了=。=