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, 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].

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        int x1 = 0;
        int x2 = x1 + 1;
        vector<int> v1;
        for(;;)
        {
            if(nums[x1] + nums[x2] == target)
            {
                v1.push_back(x1);
                v1.push_back(x2);
                return v1;
            }
            else
            {
                if(x2 == nums.size()-1)
                {
                    x1++;
                    x2 = x1+1;
                }
                else
                    x2++;
            }
        }
    }
};
备注：1 题目条件中给定了输入一定有解，而且只有唯一解，所以不需要考虑x1溢出的请客 2 因为是遍历所有情况，所以x2的最大值只会为nums.size() -1 因此不需要用》= 的条件