題目描述

Given an integer n, your task is to count how many strings of length n can be formed under the following rules:

Each character is a lower case vowel (‘a’, ‘e’, ‘i’, ‘o’, ‘u’)
Each vowel ‘a’ may only be followed by an ‘e’.
Each vowel ‘e’ may only be followed by an ‘a’ or an ‘i’.
Each vowel ‘i’ may not be followed by another ‘i’.
Each vowel ‘o’ may only be followed by an ‘i’ or a ‘u’.
Each vowel ‘u’ may only be followed by an ‘a’.
Since the answer may be too large, return it modulo 10^9 + 7.

Example 1:

Input: n = 1 Output: 5 Explanation: All possible strings are: "a", "e", "i" , "o" and "u".

Example 2:

Input: n = 2 Output: 10 Explanation: All possible strings are: "ae", "ea", "ei", "ia", "ie", "io", "iu", "oi", "ou" and "ua".

Example 3:

Input: n = 5 Output: 68

Constraints:

1 <= n <= 2 * 10^4

思路

動態規劃。dp[i][j]表示長度為i，以j為結尾，的串，個數。
根據限制條件可以推出每個字符前邊的字符是什么，以當前字符結尾的個數為所有它前面可以出現的字符的個數之和。

代碼

class Solution { public:int countVowelPermutation(int n) {int MOD = 1e9+7;vector<vector<long long int> > dp(n+1, vector<long long>(5+1, 0));for (int i=1; i<=5; ++i) {dp[1][i] = 1;}for (int i=2; i<=n; ++i) {for (int j=1; j<=5; ++j) {long long tmp = 0;if (j == 1) {tmp = (dp[i-1][2]%MOD + dp[i-1][3]%MOD + dp[i-1][5]%MOD)%MOD;}else if (j == 2) {tmp = (dp[i-1][3]%MOD + dp[i-1][1]%MOD)%MOD;}else if (j == 3) {tmp = (dp[i-1][4]%MOD + dp[i-1][2]%MOD)%MOD;}else if (j == 4) {tmp = dp[i-1][3]%MOD;}else {tmp = (dp[i-1][3]%MOD + dp[i-1][4]%MOD);}dp[i][j] += tmp;dp[i][j] %= MOD;}}long long res = 0;for (int i=1; i<=5; ++i) {res += dp[n][i];res %= MOD;}return res;} };

最后不能直接求和然后取模，會超long long。
動態規劃修行告一段落。甚至感覺比大學的時候理解的更深一點。
洗漱睡覺嘍。

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