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).
思路：和上一题差不多，将三个数的和减去target取绝对值，每次找出误差最小值及对应的三个数的和。
public class Solution {
public int threeSumClosest(int[] nums, int target) {
int len=nums.length;
int closestVal=Integer.MAX_VALUE;
int closestSum=nums[0]+nums[1]+nums[2];
Arrays.sort(nums);
int currentNum=nums[0];
for(int i=0;i<len-2;i++){
if(i>0 && nums[i]==currentNum) continue;
int begin=i+1,end=len-1;
while(begin<end){
int sum=nums[i]+nums[begin]+nums[end];
int absVal=Math.abs(sum-target);
if(absVal<closestVal){
closestVal=absVal;
closestSum=sum;
}else if(sum<target){
begin++;
}else end--;
}
currentNum=nums[i];
}
return closestSum;
}
}