Bitwise AND of Numbers Range

Description:

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.

Code:

 int rangeBitwiseAnd(int m, int n) {
        /*
        分析：经过举例发现，只要m与n之间出现2的k次方和2的k次方-1，则结果必然为0；否则，说明m一定大于等于最接近n的2的k次方。
        */
        
        //使用数学函数时注意要进行强制类型转换，否则可能出现计算结果错误的问题
        unsigned int x = pow((float)2,(int)(log((double)n)/log((double)2)));
        if ( m+1 <= x )
            return 0;
        else
        {
            //x要为无符号整型，才能得到1111111111
            x = ~0;
            for (int i = m; i <=n && i>=0; i++)
                x = x&i;
            return x;
        }
    }