[LeetCode][Java] 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).

题意：

给定一个包含n个整数的数组S,在数组中找出三个整数，使得这三个整数的和与目标值最为接近。返回这三个整数的和。你可以假定对于每个整数，都有确定的一个解。

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).

算法分析：

参考博客：http://www.zhuangjingyang.com/leetcode-3sum/

和3Sum异曲同工。 不过这里我们要判断的条件不在是三个数字和为0而是和为一个更加接近target的数字。

我们依然采用3Sum的算法，若有三个数字x1 + x2 + x3 = result 我们所求的便是让result最接近target。因此对于num，首先排序，然后遍历每个数字其下标大于自身的两个数字。然后设置两个全局变量 一个 minVal 用于记录其与target的距离，当距离减小时便更新result的新值。

AC代码：

<span style="font-size:12px;">public class Solution
{
private  int minVal = Integer.MAX_VALUE;
private  int result = 0;
public  int threeSumClosest(int[] num, int target)
{
Arrays.sort(num);
//if number is less than 3 or num is null it's can't be calc
if(num.length <3 ||  num ==null)
return target;
for(int i=0;i<num.length;i++)
{
if(i>0 && num[i] == num[i-1])
continue;
find(i,num,num[i],target);
}
return result;
}
public  void find(int index,int[] num,int target,int res)
{
int l = index+1; //low is equal to index+1 just because we just search element that is bigger than itself
int r = num.length - 1;
while(l<r)
{
if( Math.abs(num[l] + num[r] + target - res) <= minVal)
{
minVal = Math.abs(num[l] + num[r] + target - res);//it's more closer
result = num[l] + num[r] + target;
}
if(num[l] + num[r] + target >res)
r--;
else
l++;
}
}
}</span>