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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：遍历第一个数，夹逼后两个数，夹逼的前提是array必须是sorted但。记录最小的和。3sum是要求不重复组合，所以需要加条件跳过，这个并不需要。
public class Solution { public int threeSumClosest(int[] nums, int target) { if(nums==null||nums.length<3) { return 0; } Arrays.sort(nums); int size=nums.length; int min=Integer.MAX_VALUE; int save=0; for(int i=0;i<=size-3;i++) { for(int j=i+1,k=size-1;j<k;) { int sum=nums[i]+nums[j]+nums[k]; if(sum==target) { return sum; } else if(sum<target) { j++; } else { k--; } if(Math.abs(sum-target)<min) { min=Math.abs(sum-target); save=sum; } } } return save; } }
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原文地址：http://www.cnblogs.com/Machelsky/p/5940757.html
