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.
因为m!=n的话，[m, n]之间的数AND之后最低位一定是0（奇数和偶数）。然后m,n右移一位，继续计算最低位（倒数第二位）。重复上述步骤，知道范围为0，或1。用moveFactor来记录左移次数
public class Solution {
public int rangeBitwiseAnd(int m, int n) {
if(m == 0) return 0;
int moveFactor = 1;
while(m!=n){
m >>= 1;
n >>= 1;
moveFactor <<= 1;
}
return m*moveFactor;
}
}
