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1. Title
Give you an integer n, please help us to count how many strings of length n can be formed according to the following rules :
- 字符串中的每个字符都应当是小写元音字母（'a', 'e', 'i', 'o', 'u'）
- 每个元音 'a' 后面都只能跟着 'e'
- 每个元音 'e' 后面只能跟着 'a' 或者是 'i'
- 每个元音 'i' 后面 不能 再跟着另一个 'i'
- 每个元音 'o' 后面只能跟着 'i' 或者是 'u'
- 每个元音 'u' 后面只能跟着 'a'
Since the answer may be very large, so please return to mold the result after 10 ^ 9 + 7.
示例 1：
输入：n = 1
输出：5
解释：所有可能的字符串分别是："a", "e", "i" , "o" 和 "u"。
示例 2：
输入：n = 2
输出：10
解释：所有可能的字符串分别是："ae", "ea", "ei", "ia",
"ie", "io", "iu", "oi", "ou" 和 "ua"。
示例 3：
输入：n = 5
输出：68
提示：
1 <= n <= 2 * 10^4
Source: LeetCode (LeetCode)
Link: https://leetcode-cn.com/problems/count-vowels-permutation
Copyright is owned by LeetCode . For commercial reprints, please contact the official authorization. For non-commercial reprints, please indicate the source.
2. Problem solving
Similar topics:
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LeetCode 688. Probability of "horse" on the board (DP)
LeetCode 935. Knight dialer (dynamic programming)
dp[k][0-4]When the length is k, the letters are 0-4, which means the number of plans in aeiou
class Solution {
public:
int countVowelPermutation(int n) {
int mod = 1e9+7;
vector<vector<long long>> dp(n+1, vector<long long>(5, 0));
dp[1][0] = dp[1][1] = dp[1][2] = dp[1][3] = dp[1][4] = 1;
// a 0 e 1 i 2 o 3 u 4
//e,i,u --> a 上一个尾部字母 --》 下一个字母
//a, i --> e
//e o --> i
//i -- > o
//i o --> u
for(int k = 2; k <= n; ++k)
{
dp[k][0] = (dp[k-1][1]+dp[k-1][2]+dp[k-1][4])%mod;
dp[k][1] = (dp[k-1][0]+dp[k-1][2])%mod;
dp[k][2] = (dp[k-1][1]+dp[k-1][3])%mod;
dp[k][3] = (dp[k-1][2])%mod;
dp[k][4] = (dp[k-1][2]+dp[k-1][3])%mod;
}
int ans = 0;
for(int i = 0; i < 5; i++)
ans = (ans+dp[n][i])%mod;
return ans;
}
};
116 ms 27 MB
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