给定范围 [m, n]，其中 0 <= m <= n <= 2147483647，返回此范围内所有数字的按位与（包含 m, n 两端点）。
示例 1:
输入: [5,7]输出: 4
示例 2:
输入: [0,1]输出: 0
AC C++ Solution:
class Solution {public:int rangeBitwiseAnd(int m, int n) {int trans = 0;while(m != n) { //当m != n 时，移除末尾位，并记录移动位数，最后再移回来。即为范围内所有数字按位与的结果++trans;m >>= 1;n >>= 1;}return m << trans;}};
题目 [https://leetcode.com/problems/bitwise-and-of-numbers-range/][https_leetcode.com_p
（1）性质 class Solution { public: int rangeBitwiseAnd(int left, int
Given a range \[m, n\] where 0 <= m <= n <= 2147483647, return the bitwise AND of all nu
Given a range \[m, n\] where 0 <= m <= n <= 2147483647, return the bitwise AND of all nu
原题 [原题链接][Link 1] > Given a range \[m, n\] where 0 <= m <= n <= 2147483647, return t
Given a range \[m, n\] where 0 <= m <= n <= 2147483647, return the bitwise AND of all nu
> Given a range \[m, n\] where 0 <= m <= n <= 2147483647, return the bitwise AND of all
给定范围 \[m, n\]，其中 0 <= m <= n <= 2147483647，返回此范围内所有数字的按位与（包含 m, n 两端点）。 示例 1: 输入:
Given a range \[m, n\] where 0 <= m <= n <= 2147483647, return the bitwise AND of all nu
201. 数字范围按位与 201. Bitwise AND of Numbers Range 题目描述 给定范围 \[m, n\]，其中 0 <= m <= n <=
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