Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

For example, given array S = {-1 2 1 -4}, and target = 1.    The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

 

思路：

先排序，然后固定一个边界，另外两个边界收缩。

class Solution {public:    int threeSumClosest(vector
     
       &num, int target) {        int l, r, m;        int ans = num[0] + num[1] + num[2];        sort(num.begin(), num.end()); //数字从小到大排序        for(l = 0; l < num.size() - 2; l++) //固定左边界        {            m = l + 1;            r = num.size() - 1;            while(m < r) //收缩中间和右边界            {                int tmp = num[l] + num[m] + num[r];                ans = (abs(ans - target) > abs(tmp - target)) ? tmp : ans;                if(tmp < target)                {                    (num[m] > num[r]) ? r-- : m++; //比目标小 则把小的数字变大                }                else if(tmp > target)                {                    (num[m] < num[r]) ? r-- : m++; //比目标大 则把大的数字变小                }                else                {                    return tmp; //已经等于target就直接返回                }            }        }        return ans;    }};