LeetCode 201. Bitwise AND of Numbers Range

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最新推荐文章于 2024-01-19 04:58:25 发布

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分类专栏： LeetCode 题解 Multiple Passes

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本文详细介绍了如何通过位运算快速计算指定范围内整数的按位与结果，利用循环移位和计数技巧简化计算过程，特别适用于计算机科学领域的算法优化。

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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.

Suppose our given range is 5 : 0101 ---- 7: 0111

Thus, we can see that the lower 2 bits are different. By shifting them to right, we can finally make m == n. thus, result will be m << count;

Code copied from the discussion. But this method is very fantastic!!

    int rangeBitwiseAnd(int m, int n) {
        int count = 0;
        while(m != n) //until the left identical bits, all the right will be cancelled by increments from m to n;;
        {
            m >>= 1;
            n >>= 1;
            count++;
        }
        return m << count;
    }