解法一: 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;els...
Linked List Cycle II 参考资料: https://leetcode.com/problems/find-the-duplicate-number/ https://leetcode.com/problems/find-the-duplicate-number/discuss/72872/O(32*N)-solution-using-bit-manipulation-in-10-lines https://leetcode.com/problems/find-the-duplicate-number/discuss/73045/Simple-C%2B...
Although duplicate files are not complicated to deal with, it can be annoying to find and remove them one by one. If you are bothered by the large number of duplicate files in your system and want to know how to find and delete duplicate files in Windows 10 & 11. You can find the ...
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 the array (as...
Furthermore, we’ve given an intuitive explanation of the proof. The two-pointer approach is a popular technique, which can be used in many algorithmic problems. For instance, the same hare and tortoise might help us find the duplicate number in an array....
There is only one duplicate number in the array, but it could be repeated more than once. 哈希表法 复杂度 时间O(N) 空间 O(N) 思路 遍历数组时,用一个集合记录已经遍历过的数,如果集合中已经有了说明是重复。但这样要空间,不符合。 暴力法 复杂度 时间O(N^2) 空间 O(1) 思路 如果不用空间...
Your runtime complexity should be less thanO(n2). There is only one duplicate number in the array, but it could be repeated more than once 非常好的题目,开始是用二分做的,比如取数组为{1,2,3,3,4,5},mid应该是(5+1)/2 = 3,那么,如果小于等于mid的数的个数如果超过了3,那么重复的数字一...
查找环入口的方式见 LeetCode: 142. Linked List Cycle II 题解 举个栗子 求如图数组的重复元素 将数组拆分成节点, 索引作为地址, 值作为 next 指针 将节点连接成链表 整理之后,如下图: 环的入口 5, 即是重复元素 AC 代码 func findDuplicate(nums []int) int { ...
Instead of hoisting package installs in node_modules, install packages in the same manner that they are depended on. This may cause very deep directory structures and duplicate package installs as there is no de-duplicating. Sets --install-strategy=nested....
居然能由此题联想到Linked List Cycle II也是厉害. 这里的duplicate number就是cycle的起点. 课通过快慢指针找到cycle的起点. Note: 这里回置walker是回置到0. 而不是nums[0]. 因为接下来是跳动nums[runner], runner 是index. Time Complexity: O(n). Space: O(1). ...