201. Bitwise AND of Numbers Range (Bit)

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

For example, given the range [5, 7], you should return 4.

思路：序列是按1递增的，所以必定先影响低位，按位与为1的情况必定发生在高位。两个数向右移，当两数相等，剩余的1便是按位与得到的1。

class Solution {
public:
    int rangeBitwiseAnd(int m, int n) {
        int shift = 0;
        while(m!=n){
            m >>= 1;
            n >>= 1;
            shift++;
        }

        return (m << shift);
    }
};