16. 3Sum Closest.md

##16. 3Sum Closest

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

1. 排序原数列
   
2. 保存左边三个值的和在result中，移动第一个数，第二个数，第三个数的位置，再次求和，存在sum中
   
3. 比较sum - target 和 result - target 的差别，result中保存与target差值小的
   

public int threeSumClosest(int[] nums, int target) {

Arrays.sort(nums);
int result = nums[0] + nums[1] + nums[2];
  
for ( int i= 0; i<nums.length; i++){
    int j = i+1; 
    int k = nums.length -1;
   
    while (j < k){
    	int sum = nums[i] + nums[j] + nums[k]; //必须在while内
    	//比较sum， result 跟target之间的差距，result中保存跟target差距小的那个值
    	if (Math.abs(sum-target) < Math.abs(result-target))
    		result = sum; 
    	//如果sum比目标值小，左边加，否则，右边减。当sum 等于result的时候，放哪边都行。
    	if (sum < target)
    		j++;
    	else
    		k--;	
    	}	
    }
    
    return result;  
}