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(chǎn)’, ‘e’, ‘i’, ‘o’, ‘u’)
Each vowel ‘a(chǎn)’ may only be followed by an ‘e’.
Each vowel ‘e’ may only be followed by an ‘a(chǎn)’ 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(chǎn)’.
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(dòng)態(tài)規(guī)劃。dp[i][j]表示長(zhǎng)度為i，以j為結(jié)尾，的串，個(gè)數(shù)。
根據(jù)限制條件可以推出每個(gè)字符前邊的字符是什么，以當(dāng)前字符結(jié)尾的個(gè)數(shù)為所有它前面可以出現(xiàn)的字符的個(gè)數(shù)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(huì)超long long。
動(dòng)態(tài)規(guī)劃修行告一段落。甚至感覺比大學(xué)的時(shí)候理解的更深一點(diǎn)。
洗漱睡覺嘍。

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