902. Numbers At Most N Given Digit Set

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本文介绍了一个算法问题，即使用给定的数字集合构建所有可能的正整数，并统计这些整数中小于或等于指定数值N的数量。通过递归方法解决这一问题，并提供了详细的实现过程。

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We have a sorted set of digits D, a non-empty subset of {'1','2','3','4','5','6','7','8','9'}. (Note that '0' is not included.)

Now, we write numbers using these digits, using each digit as many times as we want. For example, if D = {'1','3','5'}, we may write numbers such as '13', '551', '1351315'.

Return the number of positive integers that can be written (using the digits of D) that are less than or equal to N.

Example 1:

Input: D = ["1","3","5","7"], N = 100
Output: 20
Explanation: 
The 20 numbers that can be written are:
1, 3, 5, 7, 11, 13, 15, 17, 31, 33, 35, 37, 51, 53, 55, 57, 71, 73, 75, 77.

Example 2:

Input: D = ["1","4","9"], N = 1000000000
Output: 29523
Explanation: 
We can write 3 one digit numbers, 9 two digit numbers, 27 three digit numbers,
81 four digit numbers, 243 five digit numbers, 729 six digit numbers,
2187 seven digit numbers, 6561 eight digit numbers, and 19683 nine digit numbers.
In total, this is 29523 integers that can be written using the digits of D.

Note:

D is a subset of digits '1'-'9' in sorted order.
1 <= N <= 10^9

小于n位的顺便选，等于n位的递归一下

class Solution(object):
    def atMostNGivenDigitSet(self, D, N):
        """
        :type D: List[str]
        :type N: int
        :rtype: int
        """
        def helper(s):
            n=len(s)
            if len(s)==1: return len([t for t in D if t<=s])
            res=0
            for t in D:
                if int(t)<int(s[0]):
                    res+=len(D)**(n-1)
                elif int(t)==int(s[0]):
                    res+=helper(s[1:])
            return res
            
        s=str(N)
        res=0
        for i in range(1,len(s)):
            res+=len(D)**i
        res+=helper(s)
        return res