Bitwise AND of Numbers Range LeetCode

二进制的序列有一定规律

1:  0001

2:  0010

3:  0011

4:  0100

5:  0101

6:  0110

7:  0111

8:  1000

9:  1001

10:1010

11:1011

12:1100

13:1101

14:1110

15:1111

从第0bit位（最低）起，'1'都一连续2^bitId次数出现，只要m~n不都在某一个连续的'1'区间，该bit位最后的结果必然是0

<pre name="code" class="cpp">class Solution {public:    int rangeBitwiseAnd(int m, int n) {        if (m==n) return m;        unsigned int _m=m,_n=n;        unsigned int num=n-m+1,ret=0,j,st;        for (unsigned int i=2;i<=_n;i<<=1){            if (num>i||(_m&i)==0) continue;            j=i-1;            st=_m&j;            if (i-st>=num){                ret^=i;            }        }        return ret;    }};