LeetCode 201. 数字范围按位与(Bitwise AND of Numbers Range)

201. 数字范围按位与
201. Bitwise AND of Numbers Range

题目描述
给定范围 [m, n]，其中 0 <= m <= n <= 2147483647，返回此范围内所有数字的按位与（包含 m, n 两端点）。

LeetCode201. Bitwise AND of Numbers Range中等

示例 1:
输入: [5,7]
输出: 4

示例 2:
输入: [0,1]
输出: 0

Java 实现
方法一

class Solution {
    public int rangeBitwiseAnd(int m, int n) {
        int d = Integer.MAX_VALUE;
        while ((m & d) != (n & d)) {
            d <<= 1;
        }
        return m & d;
    }
}

方法二

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

参考资料

• https://leetcode.com/problems/bitwise-and-of-numbers-range/
• https://www.cnblogs.com/grandyang/p/4431646.html
• https://leetcode-cn.com/problems/bitwise-and-of-numbers-range/

标签：201,int,Bitwise,Range,Numbers,https
来源： https://www.cnblogs.com/hglibin/p/10962434.html