201. Bitwise AND of Numbers Range

最新推荐文章于 2023-12-18 09:43:22 发布

SnailRunFast

最新推荐文章于 2023-12-18 09:43:22 发布

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本文介绍了一种高效算法，用于计算指定区间[m,n]内所有整数的按位与结果。通过寻找m和n右起第一个不同的比特位，实现了快速求解。示例输入[5,7]，输出为4；输入[0,1]，输出为0。此算法在Java在线提交中运行时间为5ms，超越了100.00%的提交。

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201. Bitwise AND of Numbers Range

Medium

Given a range [m, n] where 0 <= m <= n <= 2147483647, return the bitwise AND of all numbers in this range, inclusive.

Example 1:

Input: [5,7]
Output: 4

Example 2:

Input: [0,1]
Output: 0

Tips: find out the right most two different bits of m and n.

Runtime: 5 ms, faster than 100.00% of Java online submissions for Bitwise AND of Numbers Range.

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

        return result;
    }
}