Leetcode 16. 3Sum Closest

最新推荐文章于 2021-08-11 08:50:42 发布

sundawei2016

最新推荐文章于 2021-08-11 08:50:42 发布

阅读量181

点赞数

CC 4.0 BY-SA版权

分类专栏： Leetcode

本文链接：https://blog.youkuaiyun.com/sundawei2016/article/details/71107717

Leetcode 专栏收录该内容

83 篇文章

订阅专栏

本文介绍了一种使用双指针技术解决寻找数组中三个整数使它们的和最接近给定目标值的问题的方法。通过一次遍历及排序数组，结合双指针技巧高效求解。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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).

典型的two pointer

1、遍历一遍数组，注意 i 只能到 len-3

2、之前要算一遍result

3、然后sort array

public class Solution {
    public int threeSumClosest(int[] nums, int target) {
        int len = nums.length;
        int result = nums[0] + nums[1] + nums[len-1];
        Arrays.sort(nums);
        for (int i = 0; i < len - 2; i++) {
            int start = i + 1;
            int end = len - 1;
            while (start < end) {
                int sum = nums[i] + nums[start] + nums[end];
                if (sum > target) end--;
                else start++;
                if (Math.abs(sum - target) < Math.abs(result - target)) result = sum;
            }
        }
        return result;
    }
}