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

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给定正整数 N，返回小于等于 N 且具有至少 1 位重复数字的正整数。 

示例 1：

输入：20
输出：1
解释：具有至少 1 位重复数字的正数（<= 20）只有 11 。

示例 2：

输入：100
输出：10
解释：具有至少 1 位重复数字的正数（<= 100）有 11，22，33，44，55，66，77，88，99 和 100 。

示例 3：

输入：1000
输出：262 

提示：

1. 1 <= N <= 10^9

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Time Limit Exceeded

 1 class Solution {
 2     var N:Int = 0
 3     func numDupDigitsAtMostN(_ N: Int) -> Int {
 4         self.N = N;
 5         return N + 1 - count(0, 0)
 6         
 7     }
 8     
 9     func count(_ mask:Int,_ num:Int) -> Int
10     {
11         if num > self.N {return 0}
12         var ret:Int = 1
13         var nd:Int = num == 0 ? 1 : 0
14         while(nd < 10)
15         {
16             if ((mask>>nd) & 1) == 0
17             {
18                 ret += count(mask|(1<<nd), num*10+nd)
19             }            
20             nd += 1
21         }
22         return ret
23     }
24 }

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Time Limit Exceeded

 1 class Solution {
 2     var N:Int = 0
 3     func numDupDigitsAtMostN(_ N: Int) -> Int {
 4         var ret:Int = 0
 5         for i in 1..<10
 6         {
 7              ret += count(N, (1 << i), i)
 8         }
 9          return (N - ret)
10     }
11     
12     func count(_ limit:Int,_ mask:Int,_ num:Int) -> Int
13     {
14         if num > limit {return 0}
15         var ret:Int = 1
16         for i in 0..<10
17         {
18             if ((mask >> i) & 1) != 0{continue}
19             ret += count(limit, mask | (1 << i), num * 10 + i)
20         }
21         return ret
22     }
23 }