题目

原题在此

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

解析

一维的DP解不难, 就不多说了. 说说解析解.
这道题可以这样理解: 有5个盒子, 上面写着["a","e","i","o","u"], 有n个球放入5个盒子,盒子可以为空, 有多少种方法?
因为题目要求string为字典序,所以当组成字符串的所有字符确定了,顺序也就确定了.

盒子不为空的情况

把n个球放成一排, 插入4个隔板, 把n个小球分成5份, 板子有多少种插法?
因为盒子不为空,所以两个球之间和左右末端不能插入多个隔板. 所以有n - 1个位置插入隔板. 所以答案为 comb(n - 1 , 5 - 1)

盒子可为空的情况

先从n + 5个球中拿出5个, 在5个盒子里各放一个球, 然后再放原来的n个球. 等价于n个球放入5个盒子且盒子可以为空. 此时答案为comb(n + 5 - 1, 5 - 1)

排列组合公式

Permutation: $nP_k = \frac{n!}{(n-k)!}$
Combination: $nC_k = \frac{nP_k}{k!}$

代码

class Solution {
public:
    int countVowelStrings(int n) {
        return (n + 1) * (n + 2) * (n + 3) * (n + 4) / 24;
    }
};