leetcode201. Bitwise AND of number range

Subject: 201. Bitwise AND of Number Range

Given a range [m, n], where 0 <= m <= n <= 2147483647, return the bitwise AND of all numbers in this range (including the two ends of m, n).

Example 1:

输入: [5,7]
输出: 4

Example 2:

输入: [0,1]
输出: 0

Source: LeetCode
Link: https://leetcode-cn.com/problems/bitwise-and-of-numbers-range
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Basic idea: the longest common prefix of the boundary

• Remove the non-public part by shifting right, and count the number of digits in the public part
• Move the common part back by moving the previous result left

class Solution {
    
    
public:
    int rangeBitwiseAnd(int m, int n) {
    
    
        //这道题有一定的技巧性，直接进行位于运算，会导致一些测试用例运行超时
        //这里转化为求m,n的最长公共前缀
        int cnt = 0;
        while(m < n){
    
    
            m = m >> 1;
            n = n >> 1;
            ++cnt;
        }
        return m << cnt;
    }
};