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.

Credits:
Special thanks to @amrsaqr for adding this problem and creating all test cases.

思路：我们先以[26,30]为例，有：

11010 11011 11100 11101 11110

可见最后求得的结果是范围内所有的数公共的二进制前缀，即最终求得是最长公共二进制前缀。我们可以每次将m和n右移一位，直至m,n相等，同时记录向右移动的次数i，此时m就是我们要求的公共二进制前缀右移i位得到的数，将m左移i位即可求得最终结果。

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

201.Bitwise AND of Numbers Range

原文：http://www.cnblogs.com/zhoudayang/p/5039530.html