Leetcode: 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.

Analysis: O(N) solution will cause TLE, so this is a math problem and should generate O(1) solution

First trial: slow.

Check all 32 bits. see if both m(lower bound) and n(higher bound) are 1 on ith bit. Also need to check if diff=n-m+1 is greater than 2^i, which is the max range that 1 is fixed on this bit.

public class Solution {
    public int rangeBitwiseAnd(int m, int n) {
        int res = 0;
        int diff = n-m+1;
        int maxRange = 1;
        for (int i=0; i<=31; i++) {
            maxRange = (int)Math.pow(2, i);
            if (diff > maxRange) continue;
            int mi = (m>>i) & 1;
            int ni = (n>>i) & 1;
            if (mi == 1 && ni == 1) res |= 1<<i; 
        }
        return res;
    }
}

 

Better solution: this is actually finding the Shared Header(公共头部)

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