3Sum Closest

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).

 

第一重循环先确定第一个数字，

然后第二重循环用双指针分别从两端遍历，使得三个数的和不断接近target。

若三个数的和等于target，则可以立即返回。

时间复杂度：O(n²)

class Solution {public:    int threeSumClosest(vector
     
       &num, int target) {        sort(num.begin(), num.end());        int size = num.size();        int diff = INT_MAX;        int ret;        for(int i = 0; i < size; i ++)        {            int j = i + 1;            int k = size - 1;            while(j < k)            {                int sum = num[i] + num[j] + num[k];                if(sum == target)                    return target;                else if(sum < target)                    j ++;                else                    k --;                if(abs(sum - target) < diff)                {                    diff = abs(sum - target);                    ret = sum;                }            }        }        return ret;    }};