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>>1, n>>1) << 1) : m; } };
leetcode 201. Bitwise AND of Numbers Range(位运算,dp)
原文：http://www.cnblogs.com/zywscq/p/5429171.html