201. Bitwise AND of Numbers Range

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.

显然 m<=n， 当m < n时，最低位的And一定为0（从m到n的二进制数最低位一定存在0)，所以最低位为0；我们来看看某一位不为0的情况，就是要找到某一位没有发生二进制上的位变化的位

class Solution {
public:
    int rangeBitwiseAnd(int m, int n) {
        return (n >m)?(rangeBitwiseAnd(m >>1, n>>1) << 1):m;
    }
};

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