leetcode 16 3Sum Closest【双指针变形】

最新推荐文章于 2021-01-03 10:04:19 发布

原创最新推荐文章于 2021-01-03 10:04:19 发布 · 173 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode

leetcode 同时被 3 个专栏收录

89 篇文章

订阅专栏

LeetCode

44 篇文章

订阅专栏

python

33 篇文章

订阅专栏

本文详细解析了LeetCode第16题“最接近的三数之和”的解决方案，通过一个具体的例子[-1,2,1,-4]，目标值为1，展示了如何找到数组中三个整数相加最接近目标值的组合，并返回这三个数的和。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given an array nums of n integers and an integer target, find three integers in nums such that the sum is closest to target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

Example:

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

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

Seen this question in a real interview before?

Yes

Subscribe to see which companies asked this question.
Related Topics
Array Two Pointers

Similar Questions

和上一个LeetCode15 3sum相似

让我WA了那么就居然是因为最后一行return 缩进多写了一个？？？

class Solution(object):
    def threeSumClosest(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        nums.sort()
        ans=0x3f3f3f3f
        s=0
        for i in range(len(nums)):
            if(ans==0):
                break
            t=nums[i]
            l=i+1
            r=len(nums)-1
            while(l<r):
                if(ans>abs(target-t-nums[l]-nums[r])):
                    ans=abs(target-t-nums[l]-nums[r])
                    s=t+nums[l]+nums[r]
                if(t+nums[l]+nums[r]>target):
                    r=r-1
                elif(t+nums[l]+nums[r]<target):
                    l=l+1
                else:
                    break
        return s