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## Problem

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

You are given an integer n. There is originally an array consisting of n integers from 1 to n in ascending order, return the number of derangements it can generate. Since the answer may be huge, return it modulo $10^9 + 7$.

Example 1:

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


Example 2:

Input: n = 2
Output: 1


Constraints:

• $1 <= n <= 10^6$

## Code

Derangement: 错排问题(伯努利-欧拉装错信封问题) - 组合数学

$D(n)=(n - 1)(D(n - 2) + D(n - 1))$ $D(1) = 0; D(2) = 1$

class Solution {
public int findDerangement(int n) {
if (n == 0) return 1;
if (n == 1) return 0;

int mod = 1000000007;

long[] dp = new long[n + 1];
dp[1] = 0;
dp[2] = 1;
for (int i = 3; i <= n; i++) {
dp[i] = (dp[i - 2] + dp[i - 1]) * (i - 1) % mod;
}

return (int)dp[n];
}
}

class Solution {
public int findDerangement(int n) {
if (n == 0) return 1;
if (n == 1) return 0;

int mod = 1000000007;

long first = 0;
long second = 1;
for (int i = 3; i <= n; i++) {
long temp = second;
second = (first + second) * (i - 1) % mod;
first = temp;
}

return (int)second;
}
}