LeetCode - Count Sorted Vowel Strings

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 

Backtracking

class Solution {
    int total = 0;
    public int countVowelStrings(int n) {
        helper(n, 0, 0);
        return total;
    }
    
    public void helper(int n, int currentIndex, int currentSize) {
        if (currentSize == n) {
            total++;
            return;
        }
        for (int i = currentIndex; i < 5; i++) {
            helper(n, i, ++currentSize);
            currentSize --;
        }
    }
}

标签：Count,helper,int,++,sorted,Sorted,Input,LeetCode,currentSize
来源： https://www.cnblogs.com/incrediblechangshuo/p/14435504.html