Count Sorted Vowel Strings (M)
题目
Given an integer n, return the number of strings of length n that consist only of vowels (a, e, i, o, u) and are lexicographically sorted.
A string s is lexicographically sorted if for all valid i, s[i] is the same as or comes before s[i+1] in the alphabet.
Example 1:
Input: n = 1
Output: 5
Explanation: The 5 sorted strings that consist of vowels only are ["a","e","i","o","u"].
Example 2:
Input: n = 2
Output: 15
Explanation: The 15 sorted strings that consist of vowels only are
["aa","ae","ai","ao","au","ee","ei","eo","eu","ii","io","iu","oo","ou","uu"].
Note that "ea" is not a valid string since 'e' comes after 'a' in the alphabet.
Example 3:
Input: n = 33
Output: 66045
Constraints:
1 <= n <= 50
题意
用5个元音字母"aeiou"组成长度为n的字符串，要求排在前面的字母在字符串中同样要排在前面，求这样的字符串的个数。
思路
动态规划。用0-4来表示5个字母，dp[n][i]表示长度为n且结尾为i的字符串。可以得到递推公式：
[dp[n][i]=sum_{jle i}dp[n-1][j]
]
最后对dp[n][0-4]求和即可。
代码实现
Java
class Solution {
public int countVowelStrings(int n) {
int[][] dp = new int[n + 1][5];
for (int i = 0; i < 5; i++) {
dp[1][i] = 1;
}
for (int i = 2; i <= n; i++) {
for (int j = 0; j < 5; j++) {
for (int k = 0; k <= j; k++) {
dp[i][j] += dp[i - 1][k];
}
}
}
int sum = 0;
for (int i = 0; i < 5; i++) {
sum += dp[n][i];
}
return sum;
}
}