[LeetCode][Java] 3Sum Closest「建议收藏」

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：

给定一个包括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。你能够假定对于每一个整数。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：

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

和3Sum异曲同工。 只是这里我们要推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。

我们依旧採用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。然后遍历每一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>