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52 | 52 |
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53 | 53 | <!-- 这里可写通用的实现逻辑 -->
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54 | 54 |
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| 55 | +```bash |
| 56 | +a [e] |
| 57 | +e [a|i] |
| 58 | +i [a|e|o|u] |
| 59 | +o [i|u] |
| 60 | +u [a] |
| 61 | + |
| 62 | +=> |
| 63 | + |
| 64 | +[e|i|u] a |
| 65 | +[a|i] e |
| 66 | +[e|o] i |
| 67 | +[i] o |
| 68 | +[i|o] u |
| 69 | +``` |
| 70 | + |
| 71 | +设 `dp[i][j]` 表示当前长度为 i 且以字符 j 为结尾的字符串的数目,其中 j = {0,1,2,3,4} 分别代表元音字母 `a,e,i,o,u`。 |
| 72 | + |
55 | 73 | <!-- tabs:start -->
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56 | 74 |
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57 | 75 | ### **Python3**
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58 | 76 |
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59 | 77 | <!-- 这里可写当前语言的特殊实现逻辑 -->
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60 | 78 |
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61 | 79 | ```python
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62 |
| - |
| 80 | +class Solution: |
| 81 | + def countVowelPermutation(self, n: int) -> int: |
| 82 | + dp = (1, 1, 1, 1, 1) |
| 83 | + MOD = 1000000007 |
| 84 | + for _ in range(n - 1): |
| 85 | + dp = ((dp[1] + dp[2] + dp[4]) % MOD, (dp[0] + dp[2]) % |
| 86 | + MOD, (dp[1] + dp[3]) % MOD, dp[2], (dp[2] + dp[3]) % MOD) |
| 87 | + return sum(dp) % MOD |
63 | 88 | ```
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64 | 89 |
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65 | 90 | ### **Java**
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66 | 91 |
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67 | 92 | <!-- 这里可写当前语言的特殊实现逻辑 -->
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68 | 93 |
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69 | 94 | ```java
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| 95 | +class Solution { |
| 96 | + private static final long MOD = (long) 1e9 + 7; |
| 97 | + |
| 98 | + public int countVowelPermutation(int n) { |
| 99 | + long[] dp = new long[5]; |
| 100 | + long[] t = new long[5]; |
| 101 | + Arrays.fill(dp, 1); |
| 102 | + for (int i = 0; i < n - 1; ++i) { |
| 103 | + t[0] = (dp[1] + dp[2] + dp[4]) % MOD; |
| 104 | + t[1] = (dp[0] + dp[2]) % MOD; |
| 105 | + t[2] = (dp[1] + dp[3]) % MOD; |
| 106 | + t[3] = dp[2]; |
| 107 | + t[4] = (dp[2] + dp[3]) % MOD; |
| 108 | + System.arraycopy(t, 0, dp, 0, 5); |
| 109 | + } |
| 110 | + long ans = 0; |
| 111 | + for (int i = 0; i < 5; ++i) { |
| 112 | + ans = (ans + dp[i]) % MOD; |
| 113 | + } |
| 114 | + return (int) ans; |
| 115 | + } |
| 116 | +} |
| 117 | +``` |
| 118 | + |
| 119 | +### **C++** |
| 120 | + |
| 121 | +```cpp |
| 122 | +class Solution { |
| 123 | +public: |
| 124 | + int countVowelPermutation(int n) { |
| 125 | + using ll = long long; |
| 126 | + const ll mod = 1e9 + 7; |
| 127 | + vector<ll> dp(5, 1); |
| 128 | + vector<ll> t(5); |
| 129 | + for (int i = 0; i < n - 1; ++i) { |
| 130 | + t[0] = (dp[1] + dp[2] + dp[4]) % mod; |
| 131 | + t[1] = (dp[0] + dp[2]) % mod; |
| 132 | + t[2] = (dp[1] + dp[3]) % mod; |
| 133 | + t[3] = dp[2]; |
| 134 | + t[4] = (dp[2] + dp[3]) % mod; |
| 135 | + dp = t; |
| 136 | + } |
| 137 | + return accumulate(dp.begin(), dp.end(), 0LL) % mod; |
| 138 | + } |
| 139 | +}; |
| 140 | +``` |
70 | 141 |
|
| 142 | +### **Go** |
| 143 | +
|
| 144 | +```go |
| 145 | +func countVowelPermutation(n int) int { |
| 146 | + const mod int = 1e9 + 7 |
| 147 | + dp := [5]int{1, 1, 1, 1, 1} |
| 148 | + for i := 0; i < n-1; i++ { |
| 149 | + dp = [5]int{ |
| 150 | + (dp[1] + dp[2] + dp[4]) % mod, |
| 151 | + (dp[0] + dp[2]) % mod, |
| 152 | + (dp[1] + dp[3]) % mod, |
| 153 | + dp[2], |
| 154 | + (dp[2] + dp[3]) % mod, |
| 155 | + } |
| 156 | + } |
| 157 | + ans := 0 |
| 158 | + for _, v := range dp { |
| 159 | + ans = (ans + v) % mod |
| 160 | + } |
| 161 | + return ans |
| 162 | +} |
71 | 163 | ```
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72 | 164 |
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73 | 165 | ### **...**
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