《leetCode》：Bitwise AND of Numbers Range

题目

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.

思路一：报超时错误

思路：循环遍历按位与；

实现思路如下：

#define MAXVALUE 2147483647
int rangeBitwiseAnd(int m, int n) {
    if(0<=m&&m<=MAXVALUE&&m<=n){
        int res=m;
        for(int i=m+1;i<=n;i++){
            res&=i;
        }
        return res;
    }
    return 0;
}

思路二：寻找两个数的二进制表示时从左到右的第一个不相同的数，后面全部用零来填充

the bitwise and of the range is keeping the common bits of m and n from left to right
until the first bit that they are different, padding zeros for the rest.

#define MAXVALUE 2147483647

int rangeBitwiseAnd(int m, int n) {
    if(0<=m&&m<=MAXVALUE&&m<=n){
        int i=0;
        for(;m!=n;i++){
            m=m>>1;
            n=n>>1;
        }
        return n<<i;
    }
    return 0;
}