16. 3Sum Closest [M]

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

这个问题等于是3SUM又升级了,因为这里现在是最接近的结果,所以Hashmap的思路基本没用了,因为必须得要循环找到所有可能。当然,这个题目有个限制就是只有唯一的结果,所以,如果发现完全相等的,也就可以停下来了。

但是解题思路还是可以参考3SUM,算法基本一样,有些地方需要修改一下。首先,因为是找最接近的,所以要考虑所有情况,left循环到length-2

    for (int left = 0; left < nums.length-2; left++)

然后还是mid和right分别从两端往中央扫描,如果mid+right还比较小,那就需要mid右移,反之right左移(每次如果有最小的就存下来)

Alt text

我们可以写出如下的代码:

 mid = left+1; right = nums.length-1;
 while(mid < right)
 {
      int tmp = target-nums[left];
      if(abs(tmp - nums[mid] + nums[right]) < abs(target-Min))
          Min = nums[left] + nums[mid] + nums[right]);
      if(nums[mid] + nums[right] == tmp)
              return Min;
      else if(nums[mid] + nums[right] < tmp)
              mid++;
      else
             right--;
 }

代码

public class Solution {
    public int threeSumClosest(int[] nums, int target) {       
        Arrays.sort(nums);
        int mid,right;
        if(nums.length < 3)
            return 0;
        int Min = nums[0]+nums[1]+nums[2];
        //left要循环全部
        for (int left = 0; left < nums.length-2; left++) {
            mid = left+1; right = nums.length-1;
            int tmp = target-nums[left];
            while(mid < right)
            {
                if(Math.abs(tmp - nums[mid] - nums[right]) <Math.abs(target - Min))  //每次查看是不是最小的情况
                    Min = nums[left]+ nums[mid] + nums[right];
                if(nums[mid] + nums[right] == tmp)
                {
                    return target;   //因为只有一种答案所以可以直接返回
                }
                else if(nums[mid] + nums[right] < tmp)
                    mid++;
                else
                    right--;
            }
        }
        return Min;
    }
}

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