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.

题解:如果m==n,那么答案就是m.

如果m<n,那么二进制最右边一位在最后的结果中肯定是0,那么就可以转化成子问题:

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

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