Create 201._Bitwise_AND_of_Numbers_Range.md

docs/Leetcode_Solutions/201._Bitwise_AND_of_Numbers_Range.md

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+# 201. Bitwise AND of Numbers Range
+
+**<font color=red>难度: Medium</font>**
+
+## 刷题内容
+
+> 原题连接
+
+* https://leetcode.com/problems/bitwise-and-of-numbers-range/description/
+
+> 内容描述
+
+```
+Given a range [m, n] where 0 <= m <= n <= 2147483647, return the bitwise AND of all numbers in this range, inclusive.
+
+Example 1:
+
+Input: [5,7]
+Output: 4
+Example 2:
+
+Input: [0,1]
+Output: 0
+```
+
+## 解题方案
+
+> 思路 1
+******- 时间复杂度: O(1)******- 空间复杂度: O(1)******
+
+我们知道你从m一直&到n，那么m和n二进制位后面不一样的地方肯定都会变成0
+
+所以我们找到m和n的最长相同前缀，然后补齐二进制的32位（即往右边贴'0'）
+
+自己写的一行版本！
+
+
+```python
+import os
+class Solution(object):
+    def rangeBitwiseAnd(self, m, n):
+        """
+        :type m: int
+        :type n: int
+        :rtype: int
+        """
+        return int(os.path.commonprefix([bin(m)[2:].zfill(32), bin(n)[2:].zfill(32)])[::-1].zfill(32)[::-1], 2)
+```
+
+
+> 思路 2
+******- 时间复杂度: O(1)******- 空间复杂度: O(1)******
+
+
+别人的一行版本！
+
+
+```python
+class Solution(object):
+    def rangeBitwiseAnd(self, m, n):
+        """
+        :type m: int
+        :type n: int
+        :rtype: int
+        """
+        return self.rangeBitwiseAnd(m, n & n-1) if m < n else n
+```