[LeetCode] Find the Derangement of An Array 找数组的错排

In combinatorial mathematics, a derangement is a permutation of the elements of a set, such that no element appears in its original position.

There‘s originally an array consisting of n integers from 1 to n in ascending order, you need to find the number of derangement it can generate.

Also, since the answer may be very large, you should return the output mod 109 + 7.

Example 1:

Input: 3
Output: 2
Explanation: The original array is [1,2,3]. The two derangements are [2,3,1] and [3,1,2].

Note:
n is in the range of [1, 106].

时间： 2024-08-06 08:19:31

[LeetCode] Find the Derangement of An Array 找数组的错排的相关文章

[LeetCode] Find All Duplicates in an Array 找出数组中所有重复项

Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all the elements that appear twice in this array. Could you do it without extra space and in O(n) runtime? Example: Input: [4,3,2,7,

LeetCode 977. Squares of a Sorted Array (有序数组的平方)

题目标签:Array 题目给了我们一组从小到大的 integers,让我们平方数字并且也排序成从小到达. 因为有负数在里面,平方后,负数在array的位置会变动. 可以设left 和 right pointers,从两边遍历,比较一下两个平方后的数字,把大的那个放入新建的array的末尾. 具体看code. Java Solution: Runtime beats 100.00% 完成日期:02/03/2019 关键点:two pointers 1 class Solution 2 {

leetCode 26.Remove Duplicates from Sorted Array(删除数组重复点) 解题思路和方法

Remove Duplicates from Sorted Array Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length. Do not allocate extra space for another array, you must do this in place with constant memory.

[leetcode] 80 Remove Duplicates from Sorted Array II(数组下标操作)

因为这道题目的题意是要求我们在原数组上进行操作,所以操作变得稍微复杂了些,否则直接使用map最为简单. 基本思想是记录两个指针,一个是当前数组,另一个是目的数组,注意如果发现重复数超过2,那么目的数组的cur就要阻塞, 直到不同的出现后再赋值前进. class Solution { public: int removeDuplicates(vector<int>& nums) { if(nums.size()==0) return 0; int cur=1; //修改后数组的下标点 i

[LeetCode] 033. Search in Rotated Sorted Array (Hard) (C++)

索引:[LeetCode] Leetcode 题解索引 (C++/Java/Python/Sql) Github: https://github.com/illuz/leetcode 033. Search in Rotated Sorted Array (Hard) 链接: 题目:https://leetcode.com/problems/search-in-rotated-sorted-array/ 代码(github):https://github.com/illuz/leetcode 题

LeetCode 80 Remove Duplicates from Sorted Array II [Array/auto] <c++>

LeetCode 80 Remove Duplicates from Sorted Array II [Array/auto] <c++> 给出排序好的一维数组,如果一个元素重复出现的次数大于两次,删除多余的复制,返回删除后数组长度,要求不另开内存空间. C++ 献上自己丑陋无比的代码.相当于自己实现一个带计数器的unique函数 class Solution { public: int removeDuplicates(std::vector<int>& nums) {

【LeetCode】Remove Duplicates from Sorted Array 解题报告

[LeetCode]Remove Duplicates from Sorted Array 解题报告标签(空格分隔): LeetCode [LeetCode] https://leetcode.com/problems/remove-duplicates-from-sorted-array/ Total Accepted: 129010 Total Submissions: 384622 Difficulty: Easy Question Given a sorted array, remov

Java for LeetCode 081 Search in Rotated Sorted Array II

Follow up for "Search in Rotated Sorted Array": What if duplicates are allowed? Would this affect the run-time complexity? How and why? Write a function to determine if a given target is in the array. 解题思路: 参考Java for LeetCode 033 Search in Rota