#16 3Sum Closest

题目如下：

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).

代码如下：

public class Solution {    public int threeSumClosest(int[] nums, int target) {        Arrays.sort(nums);         int len=nums.length;        int closet=nums[0]+nums[1]+nums[2];        for(int i=0;i<len-2;i++){            int l=i+1;//left            int r=len-1;//right            /*if(l==r-1){                closet=Math.abs(closet-target)<Math.abs(nums[i]+nums[l]+nums[r]-target)?closet:nums[i]+nums[l]+nums[r];                continue;            }*/            while(l<r-1){                if(closet==target)return closet;                int a=nums[i]+nums[l]+nums[r]-target;                if(nums[l]+nums[r]<target-nums[i])                    l++;                else r--;                int b=nums[i]+nums[l]+nums[r]-target;                if(a*b<0){                    int min=Math.abs(a)<Math.abs(b)?a:b;                    closet=Math.abs(closet-target)<Math.abs(min)?closet:min+target;                    continue;                }            }            closet=Math.abs(closet-target)<Math.abs(nums[i]+nums[l]+nums[r]-target)?closet:nums[i]+nums[l]+nums[r];                    }        return closet;    }}

public class Solution {public int threeSumClosest(int[] num, int target) {int min = Integer.MAX_VALUE;int result = 0; Arrays.sort(num); for (int i = 0; i < num.length; i++) {int j = i + 1;int k = num.length - 1;while (j < k) {int sum = num[i] + num[j] + num[k];int diff = Math.abs(sum - target); if(diff == 0) return sum; if (diff < min) {min = diff;result = sum;}if (sum <= target) {j++;} else {k--;}}} return result;}}