描述
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
Note:
1 <= n <= 50
解析
根据题意，只要先找出规律就可以用动态规划的思想去做，首先找规律，当 n 为 1 的时候，以每种元音开头的合法字符串有多少种，可以这样如下列出：
a e i o u
1 1 1 1 1
当 n 为 2 的时候，可以如下列出：
a e i o u
1 1 1 1 1
5 4 3 2 1
当 n 为 3 的时候，可以如下列出：
a e i o u
1 1 1 1 1
5 4 3 2 1
15 10 6 3 1
以此类推，其中的规律很明显，初始化一个数组 res ，res 有 5 个元素，没个元素对应的是分别以 5 个元音开头的字符串可以出现的合法次数，res[1] 表示的是以 a 开头的合法字符粗汉，以此类推。遍历 n 次，每次遍历的时候，只需要将 res[j] 变为 sum(res[j:]) 即可，遍历结束求 sum(res) 即可得到答案。
解答
class Solution(object):
def countVowelStrings(self, n):
"""
:type n: int
:rtype: int
"""
res = [1,1,1,1,1]
for i in range(1, n):
for j in range(5):
res[j] = sum(res[j:])
return sum(res)
运行结果
Runtime: 16 ms, faster than 86.61% of Python online submissions for Count Sorted Vowel Strings.
Memory Usage: 13.3 MB, less than 93.61% of Python online submissions for Count Sorted Vowel Strings.
解析
也可以使用 DFS 来解决， n 就相当于 DFS 的递归深度，宽度就是 5 ，因为是 5 个元音字母之间去排列组合，但是因为递归栈的深度太深，导致耗时严重，刚刚可以 AC ，这时就更显出上面动态规划的省时省力了。
解答
class Solution(object):
def __init__(self):
self.result = 0
def countVowelStrings(self, n):
"""
:type n: int
:rtype: int
"""
def dfs(idx, length, L, vowels):
if length == n:
self.result += 1
return
for i in range(idx, len(vowels)):
dfs(i, length + 1, L + vowels[i], vowels)
dfs(0, 0, '', 'aeiou')
return self.result
运行结果
Runtime: 7964 ms, faster than 5.56% of Python online submissions for Count Sorted Vowel Strings.
Memory Usage: 13.3 MB, less than 83.33% of Python online submissions for Count Sorted Vowel Strings.
