Create 1015._Numbers_With_1_Repeated_Digit.md

docs/Leetcode_Solutions/Python/1015._Numbers_With_1_Repeated_Digit.md

+# 1015. Numbers With 1 Repeated Digit
+
+**<font color=red>难度: Hard</font>**
+
+## 刷题内容
+
+> 原题连接
+
+* https://leetcode.com/problems/numbers-with-1-repeated-digit/
+
+> 内容描述
+
+```
+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 <= N <= 10^9
+```
+
+## 解题方案
+
+> 思路 1
+******- 时间复杂度: O(lgN)******- 空间复杂度: O(lgN)******
+
+算有重复的，我们可以先算出没有重复的，然后返回`(总数-没有重复)`即可
+
+那么没有重复怎么算呢？
+
+对于一个N位digit的数字，小于等于它的没有重复的有以下两种：
+1. digit数小于N的，对于这部分，直接算即可
+2. digit数小于N的，对于这部分，我们必须要让构造出的数字小于等于N
+
+
+```python
+class Solution:
+    def numDupDigitsAtMostN(self, N):
+        L = list(map(int, str(N + 1)))
+        res, n = 0, len(L)
+
+        def A(m, n):
+            return 1 if n == 0 else A(m, n - 1) * (m - n + 1)
+
+        for i in range(1, n): 
+            res += 9 * A(9, i - 1)
+        s = set()
+        for i, x in enumerate(L):
+            for y in range(0 if i else 1, x):
+                if y not in s:
+                    res += A(9 - i, n - i - 1)
+            if x in s: 
+                break
+            s.add(x)
+        return N - res
+```
+
+```
+For anyone who doesn't understand the function `def A()`, you can see the iterative version of it below.
+```python
+        def A(m, n):
+            res = 1
+            for i in range(n):
+                res *= m
+                m -= 1
+            return res
+```
+It means the `permutation` of `m * (m-1) * ... * (m-(n-1))`.
+
+And for why use `9*A(9,i-1)` instead of `A(9,i)`, that's because there should be no leading `0`, but the following digit can be `0`.
+```
+
+***将函数A重命名为permu***
+
+```python
+class Solution:
+    def numDupDigitsAtMostN(self, N):
+        L = list(map(int, str(N + 1)))
+        res, n = 0, len(L)
+
+        def permu(m, n):
+            res = 1
+            for i in range(n):
+                res *= m
+                m -= 1
+            return res
+
+        for i in range(1, n): 
+            res += 9 * permu(9, i - 1)
+            
+        s = set()
+        for i, x in enumerate(L):
+            for y in range(0 if i else 1, x):
+                if y not in s:
+                    res += permu(9 - i, n - i - 1)
+            if x in s: 
+                break
+            s.add(x)
+        return N - res
+```
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