LeetCode 16.3Sum Closest

最新推荐文章于 2018-11-19 17:53:00 发布

茶生

最新推荐文章于 2018-11-19 17:53:00 发布

阅读量289

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode-P 文章标签： leetcode python

本文链接：https://blog.youkuaiyun.com/Lu_gee/article/details/75016395

LeetCode-P 专栏收录该内容

22 篇文章

订阅专栏

本文介绍了一个经典的算法问题——寻找数组中三个整数，使其和最接近给定的目标值，并提供了一种有效的解决方案。通过先排序再使用双指针技术遍历数组，能够高效地找到满足条件的三个数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目描述：

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).

AC代码：

class Solution(object):
    def threeSumClosest(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        nums.sort()
        result = []
        for i in range(len(nums) - 2):
            l, r = i + 1, len(nums) - 1
            while l < r:
                sums = nums[i] + nums[l] + nums[r]
                diff = sums - target
                if diff == 0:
                    return target
                elif diff < 0:
                    result.append(diff)
                    l += 1
                else:
                    result.append(diff)
                    r -= 1
        result.sort()
        if result[0] > 0:
            return target + result[0]
        elif result[-1] < 0:
            return target + result[-1]
        else:
            for i in range(len(result)):
                if result[i] < 0 < result[i + 1]:
                    if abs(result[i]) < result[i + 1]:
                        return target + result[i]
                    else:
                        return target + result[i + 1]