LeetCode OJ算法题（十六）：3Sum Closest

题目：

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).

解法：

与上一篇博客3Sum解法遍历方法相同，唯一需要改变的是在每一次Xi+Xj时，都比较三数和与target的误差，如果误差减小，则更新误差与返回值ret。

不再赘述，直接上代码

import java.util.Arrays;public class No16_ThreeSumClosest {public static void main(String[] args){System.out.println(threeSumClosest(new int[]{0,1,2}, 0));}public static int threeSumClosest(int[] num, int target){if(num.length < 3) return 0;Arrays.sort(num);int ret = num[0] + num[1] + num[num.length-1];int err = ret - target;if(err < 0) err *= -1;int k_pre = num[num.length-1]+1;for(int k=0;k<num.length-2;k++){if(num[k] == k_pre) continue;k_pre = num[k];int i = k+1;int j = num.length-1;while(i<j){if(num[i]+num[j] < target-num[k]){if(err > target-num[k]-num[i]-num[j]){err = target-num[k]-num[i]-num[j];ret = num[k]+num[i]+num[j];}i++;continue;}if(num[i]+num[j] > target-num[k]){if(err > num[i]+num[j]+num[k]-target){err = num[i]+num[j]+num[k]-target;ret = num[k]+num[i]+num[j];}j--;continue;}return target;}}        return ret;}}