GuilinDev

Lc0016

05 August 2008

16 - 3Sum Closest

原题概述

Given an array 

1

nums
of n integers and an integer 
1

target
, find three integers in 
1

nums
such that the sum is closest to 
1

target
. Return the sum of the three integers. You may assume that each input would have exactly one solution.

Example:

1
2
3

Given array nums = [-1, 2, 1, -4], and target = 1.

The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

题意和分析

找到数组中的一个triplet三个数的和距离target的差值相比其它triplets来说是最小的，并返回这个triplet的和，因为不是返回所有可能的triplets，所以不需要去重了。这是上一道题的延伸，思路类似，先排序，然后确定一个index，剩下的两个indices两头扫描。

同样，时间复杂度: O(nlogn) + O(n^2) = O(n^2)；空间复杂度O(n)。

代码

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class Solution {
    public int threeSumClosest(int[] nums, int target) {
        if (nums == null || nums.length < 3) {
            return 0;
        }
        
        int result = 0; 
        int min = Integer.MAX_VALUE;
        
        Arrays.sort(nums);
        
        for (int first = 0; first <= nums.length - 3; first++) {
            int second = first + 1, third = nums.length - 1;
            while (second < third) {//循环一轮找到最小的min
                int sum = nums[first] + nums[second] + nums[third];
                if (Math.abs(sum - target) < min) {
                    min = Math.abs(sum - target);
                    result = sum;
                }
                
                if (sum == target) {//等于target那就是差距最小了，直接返回
                    return sum;
                } else if (sum < target) {
                    second++;
                } else {
                    third--;
                }
            }
        }
        return result;
    }
}
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