Find the Duplicate Number LeetCode:287. Find the Duplicate Number 找到数组中重复多次的那个数(唯一)。 思路一:元素值作索引 从位置0开始遍历数组,每次得到的元素值作为下一个遍历的元素,并且将遍历过的元素置为0,如果存在重复的元素值,那么这个位置的元素肯定会被多次访问,如果访问到的元素值为0则表明这个...
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...
public class Solution { public int findDuplicate(int[] nums) { int slow = 0; int fast = 0; // 找到快慢指针相遇的地方 do{ slow = nums[slow]; fast = nums[nums[fast]]; } while(slow != fast); int find = 0; // 用一个新指针从头开始,直到和慢指针相遇 while(find != slow){ slo...
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...
To see the algorithm in action, check out the animation from:https://leetcode.com/problems/find-the-duplicate-number/solution/ classSolution:deffindDuplicate(self, nums):#Find the intersection point of the two runners.tortoise =nums[0]
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...
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, …
public class Solution { public int findDuplicate(int[] nums) { int slow = 0; int fast = 0; // 找到快慢指针相遇的地方 do{ slow = nums[slow]; fast = nums[nums[fast]]; } while(slow != fast); int find = 0; // 用一个新指针从头开始,直到和慢指针相遇 ...
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
fast=nums[nums[fast]];if(slow == fast)break; }while(true) { slow=nums[slow]; t=nums[t];if(slow == t)break; }returnslow; } }; 本文转自博客园Grandyang的博客,原文链接:寻找重复数[LeetCode] Find the Duplicate Number,如需转载请自行联系原博主。