問題描述：
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].

解法一：
拿到題目首先就想到了暴力遍歷，雙重迴圈時間複雜度O(n^2)，是最耗時的演算法。

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> tar;
        for(int i=0; i<nums.size(); i++){
            for(int j=i+1; j<nums.size(); j++){
                if(nums[j] == target - nums[i])
                {
                    tar.push_back(i);
                    tar.push_back(j);
                    break;
                }
            }
        }
        return tar;
    }
};
 

其他解法之後更新！