Given a range [m, n] where 0 <= m <= n <= 2147483647, return the bitwise AND of all numbers in this range, inclusive.
For example, given the range [5, 7], you should return 4.
class Solution {public:int rangeBitwiseAnd(int m, int n) {int re = 0;if (m ==n)return m;int k1 = log2(m);int k2 = log2(n);while (k2== k1){re += pow(2, k1);m = m - pow(2, k1);n = n - pow(2, k1);k1 = log2(m);k2 = log2(n);}return re;}};
accepted
题目 [https://leetcode.com/problems/bitwise-and-of-numbers-range/][https_leetcode.com_p
Given a range \[m, n\] where 0 <= m <= n <= 2147483647, return the bitwise AND of all nu
这道题是leetcode上新出的题目，有兴趣做了一下，轻松通过，具体代码如下所示 class Solution { public: int r
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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