【LeetCode】201. Bitwise AND of Numbers Range

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.

从m到n逐个做与操作肯定是不合理的，最多要进行INT_MAX*32次位与操作。

可以把复杂度降低到32次移位并处理。

对于每一位来说，只要中间存在一个0，该位就是0，只有全1的时候才是1.

因此问题变为：对于从右数第i位数字，从m到n之间是否全1？

满足全1要同时满足两个条件：

（1）m的相应位置是1 即起始位置必须是1

（2）m到n之间的间隔不大于m到最后一个连续1的间隔 

class Solution {
public:
    int rangeBitwiseAnd(int m, int n) {
        int ret = 0;
        int gap = n - m;
        if(gap == 0)
            return m;
        int bit = 1;
        //note that 0 <= m <= n <= 2147483647
        //the highest bit must be 0, thus skip i == 31
        for(int i = 0; i < 31; i ++)
        {//bit by bit check zero
            int ind = m % (int)pow(2.0, i+1);
            if((ind >= (int)pow(2.0, i)) && (ind+gap <= (int)pow(2.0, i+1)-1))
            //all 1's
                ret |= bit;
            bit <<= 1;
        }
        return ret;
    }
};