leetcode 16. 3Sum Closest

最新推荐文章于 2020-10-13 04:19:10 发布

原创最新推荐文章于 2020-10-13 04:19:10 发布 · 263 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode

LeetCode 专栏收录该内容

41 篇文章

订阅专栏

本文介绍了一个算法问题：在整数数组中找到三个数，使其和最接近给定的目标值。通过排序和双指针技巧，实现了一个高效的解决方案。

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).

在一组数组中找出一个三元组(a,b,c)的和,其和最接近target.

和leetcode 15.3Sum题没多大区别,一样的思路.

public class A16_3SumClosest {
	public int threeSumClosest(int[] nums, int target) {
		Arrays.sort(nums);
		int distance = Integer.MAX_VALUE; // 3元组的和与target的距离
		int ans = nums[0];
		for(int i = 0; i < nums.length - 2; i++) {
			int second = i + 1;
			int third = nums.length - 1;
			while(second < third) {
				int sum = nums[i] + nums[second] + nums[third];
				if(sum == target) {
					return target;
				}
				if (Math.abs(sum - target) < distance) {
					distance = Math.abs(sum - target);
					ans = sum;
				}
				if(sum > target) {
					do {
						third--;
					} while(second < third && nums[third + 1] == nums[third]);
				} else {
					do {
						second++;
					} while(second < third && nums[second - 1] == nums[second]);
				}
			}
			while(i < nums.length - 2 && nums[i] == nums[i + 1])
				i++;
		}
		return ans;
    }
}