1015. Numbers With 1 Repeated Digit

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Given a positive integer N, return the number of positive integers less than or equal to N that have at least 1 repeated digit.

Example 1:

Input: 20
Output: 1
Explanation: The only positive number (<= 20) with at least 1 repeated digit is 11.

Example 2:

Input: 100
Output: 10
Explanation: The positive numbers (<= 100) with atleast 1 repeated digit are 11, 22, 33, 44, 55, 66, 77, 88, 99, and 100.

Example 3:

Input: 1000
Output: 262

Note:

1. 1 <= N <= 10^9

思路：求出没有重复的总数A，然后N-A就是结果。求A可以用DFS，开一个数组ps表示前面用了什么数字，要注意首位为0的情况（最烦这种题了...）

class Solution(object):
    def numDupDigitsAtMostN(self, N):
        """
        :type N: int
        :rtype: int
        """
        def fac(n,cnt):
            res=1
            for i in range(n,n-cnt,-1): res*=i
            return res
        
        def all_diff(n,idx,ps):
            if idx==len(str(n)): return 1
            
            res = 0
            dig = int(str(n)[idx])
            for i in range(dig):
                if i in ps: continue
                if idx==0 and i==0:
                    for j in range(i+1,len(str(n))):
                        res+=9*fac(9,len(str(n))-j-1)
                else:
                    res+=fac(10-len(ps)-1,len(str(n))-idx-1)
            
            if dig not in ps:
                ps.append(dig)
                res+=all_diff(n,idx+1,ps)
                ps.pop()
            
            return res
        
        t = all_diff(N,0,[])
#        print(t)
        return N-t