Leetcode 287. Find the Duplicate Number Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one. Note: You must not modify...
Given an arraynumscontainingn+ 1 integers where each integer is between 1 andn(inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one. Note: You must not modify the array (assume the array is read only). Yo...
[Leetcode]287. Find the Duplicate Number 这是Leetcode第287题,给定一个包含n + 1个整数的数组,其中每一个整数均介于[1, n]之间,证明其中至少有一个重复元素存在。假设只有一个数字出现重复,找出这个重复的数字。 这道题可以用计数排序的思路求解,但更为巧妙的方法是使用快慢指针求解。 基本思想是将数组抽象...
Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, …
题目链接: Find the Duplicate Number : leetcode.com/problems/f 寻找重复数: leetcode-cn.com/problem LeetCode 日更第 73 天,感谢阅读至此的你 欢迎点赞、收藏、在看鼓励支持小满 发布于 2022-03-30 07:57 力扣(LeetCode) 算法 二分查找 赞同1添加评论 分享喜欢收藏申请转载...
LeetCode 287. Find the Duplicate Number 简介:给定一个包含 n + 1 个整数的数组 nums,其数字都在 1 到 n 之间(包括 1 和 n),可知至少存在一个重复的整数。假设只有一个重复的整数,找出这个重复的数。 Description Given an array nums containing n + 1 integers where each integer is between 1 and...
Leetcode每日一题:287.find-the-duplicate-number(寻找重复数),思路:一开始并没有什么头绪,直接排序加遍历以O(nlgn)的复杂度水过去了,后来看评论才知道有Floyd判圈算法这么秒的方法,简称龟兔赛跑;具体算法讲解可参考文章:算法-floyd判环(圈)算法,讲得很棒,便于理
LeetCode: 287. Find the Duplicate Number 题目描述 Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one. ...
287. 寻找重复数 - 给定一个包含 n + 1 个整数的数组 nums ,其数字都在 [1, n] 范围内(包括 1 和 n),可知至少存在一个重复的整数。 假设 nums 只有 一个重复的整数 ,返回 这个重复的数 。 你设计的解决方案必须 不修改 数组 nums 且只用常量级 O(1) 的额外空间。
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