Two Sum 找出数组内两个数相加和为n的下标

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].

 看到个有趣的算法，也许时间复杂度不是最间的，但这个减而治之还是值得借鉴。

通过排序后，最小值+最大值比较，如果两数相加大于目标值，则最大值序号减一，最小值+次大值。反之，次小值加最大值。

class Solution {
public:
    struct S{
        int val,pos;
        friend bool operator < (S &s1,S &s2){
            return s1.val<s2.val;
        }
    };
    vector<int> twoSum(vector<int>& nums, int target) {
        int len=nums.size();
        struct S s[len];
        for(int i=0;i<len;++i){
        	s[i].val=nums[i];
        	s[i].pos=i;
        }
        sort(s,s+len);
        vector<int> ans;
        int h=0,r=len-1;
        if(s[r].val+s[r-1].val<target||s[h].val+s[h+1].val>target) return ans;
        while(h<r){
        	if(s[h].val+s[r].val==target){
        		ans.push_back(s[h].pos);
        		ans.push_back(s[r].pos);
        		break;
        	}else if(s[h].val+s[r].val<target){
        		++h;
        	}else{
        		--r;
        	}
        }
        return ans;
    }
};