201. Bitwise AND of Numbers Range 解题记录

题目描述：

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.

解题思路：

通过观察可以得出，对m和n之间（包括m, n）的所有数字进行&操作，就相当于找到m, n二进制中左边相同的1。比如：

5    101
6    110
7    111

4    100

 

代码：

class Solution {
public:
    int rangeBitwiseAnd(int m, int n) {
        int bits = 0;  //右移的位数，最后再左移回来
        while(m != n && m != 0){
            m = m>>1;
            n = n>>1;
            bits++;
        }
        return m<<bits;
    }
};