类似LeetCode 442. Find All Duplicates in an Array,由于元素是1~n,因此每个元素的值-1(映射到0~n-1)就可以直接当做下标。 classSolution {public:intfindDuplicate(vector<int>&nums) {for(inti=0;i<nums.size();++i){intindex=abs(nums[i])-1;if(nums[index]>0) nums[index]*=-1;elsereturnind...
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...
classSolution:deffindDuplicate(self, nums:List[int]) ->int:# 如果只有两个元素,第一个元素一定是重复元素iflen(nums) ==2:returnnums[0]# fast每次走两步,slow每次走一步,起始点可以为任意位置fast =0slow =0# python没有do while,所以在循环外写了一遍slow = nums[slow] fast = nums[nums[fast]]...
[Leetcode] Find the Duplicate Number 找到重复数字 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 ...
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, …
LeetCode:287. 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...
代码语言:javascript 代码运行次数:0 classSolution{public:intfindDuplicate(vector<int>&nums){if(nums.size()>1){int slow=nums[0];int fast=nums[nums[0]];while(slow!=fast){slow=nums[slow];fast=nums[nums[fast]];}fast=0;while(fast!=slow){fast=nums[fast];slow=nums[slow];}returnslow;}re...
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 ...
https://leetcode.com/problems/find-the-duplicate-number/ 题目: nums containing n + 1 integers where each integer is between 1 and n Note: You must not You must use only constant, O(1) extra space. Your runtime complexity should be less than ...
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