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
题目大意：
给出一个区间，我们需要求出其中全部的数进行或运算之后的结果。
解题思路：
我们发现如果某一列中存在一个0即该处的最终结果为0，所以我们需要找出这些数前部分的公共部分。
再将这个数右移。
class Solution {public:int rangeBitwiseAnd(int m, int n) {int step = 0;while(m!=n){m>>=1;n>>=1;step++;}return n<<step;}};
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
一：[Number of 1 Bits][] 题目： Write a function that takes an unsigned integer and returns
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 <=
偏执的太偏执、
àì夳堔傛蜴生んèń
傷城~
我就是我
爱被打了一巴掌
深藏阁楼爱情的钟
还没有评论，来说两句吧...