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‎README.adoc

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//|{counter:codes}
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//|{leetcode_base_url}/bitwise-and-of-numbers-range/[201. Bitwise AND of Numbers Range^]
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//|{source_base_url}/_0201_BitwiseANDOfNumbersRange.java[Java]
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//|{doc_base_url}/0201-bitwise-and-of-numbers-range.adoc[题解]
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//|Medium
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//|
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|{counter:codes}
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|{leetcode_base_url}/bitwise-and-of-numbers-range/[201. Bitwise AND of Numbers Range^]
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|{source_base_url}/_0201_BitwiseANDOfNumbersRange.java[Java]
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|{doc_base_url}/0201-bitwise-and-of-numbers-range.adoc[题解]
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|{counter:codes}
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|{leetcode_base_url}/happy-number/[202. Happy Number^]

‎docs/0201-bitwise-and-of-numbers-range.adoc

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[#0201-bitwise-and-of-numbers-range]
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= 201. Bitwise AND of Numbers Range
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= 201. 数字范围按位与
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{leetcode}/problems/bitwise-and-of-numbers-range/[LeetCode - Bitwise AND of Numbers Range^]
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https://leetcode.cn/problems/bitwise-and-of-numbers-range/[LeetCode - 201. 数字范围按位与 ^]
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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.
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给你两个整数 `left` 和 `right` ，表示区间 `[left, right]`，返回此区间内所有数字 *按位与* 的结果（包含 `left` 、`right` 端点）。
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*Example 1:*
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*示例 1：*
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[subs="verbatim,quotes,macros"]
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----
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*Input:* [5,7]
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*Output:* 4
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....
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输入：left = 5, right = 7
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输出：4
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....
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----
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*示例 2：*
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*Example 2:*
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....
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输入：left = 0, right = 0
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输出：0
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....
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[subs="verbatim,quotes,macros"]
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----
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*Input:* [0,1]
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*Output:* 0
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----
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*示例 3：*
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....
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输入：left = 1, right = 2147483647
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输出：0
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....
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*提示：*
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* `0 \<= left \<= right \<= 2^31^ - 1`
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== 思路分析
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当计算 stem:[n & (n - 1)] 时，会把 `n` 最右边的 `1` 编程 `0`。
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image::images/0201-10.png[{image_attr}]
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image::images/0201-11.png[{image_attr}]
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从另外一个角度来看，就是寻找指定范围内所有数字的公共前缀。
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image::images/0201-12.png[{image_attr}]
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[[src-0201]]
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[tabs]
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====
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一刷::
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--
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[{java_src_attr}]
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----
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include::{sourcedir}/_0201_BitwiseANDOfNumbersRange.java[tag=answer]
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include::{sourcedir}/_0201_BitwiseAndOfNumbersRange.java[tag=answer]
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----
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--
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// 二刷::
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// +
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// --
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// [{java_src_attr}]
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// ----
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// include::{sourcedir}/_0201_BitwiseAndOfNumbersRange_2.java[tag=answer]
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// ----
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// --
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====
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== 参考资料
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. https://leetcode.cn/problems/bitwise-and-of-numbers-range/solutions/384938/shu-zi-fan-wei-an-wei-yu-by-leetcode-solution/[201. 数字范围按位与 - 官方题解^]
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. https://leetcode.cn/problems/bitwise-and-of-numbers-range/solutions/385466/ju-hao-li-jie-de-wei-yun-suan-si-lu-by-time-limit/[201. 数字范围按位与 - 巨好理解的位运算思路^]
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. https://leetcode.cn/problems/bitwise-and-of-numbers-range/solutions/538550/golang-yi-xing-suan-fa-by-endlesscheng-iw6y/[201. 数字范围按位与 - O(1) 写法^]
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. https://leetcode.cn/problems/bitwise-and-of-numbers-range/solutions/56903/xiang-xi-tong-su-de-si-lu-fen-xi-duo-jie-fa-by--41/[201. 数字范围按位与 - 详细通俗的思路分析，多解法^]
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‎docs/images/0201-10.png

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‎docs/images/0201-11.png

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‎docs/images/0201-12.png

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‎docs/index.adoc

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‎logbook/202503.adoc

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|{doc_base_url}/0173-binary-search-tree-iterator.adoc[题解]
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|✅ 栈。使用栈模拟递归调用。
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|{counter:codes2503}
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|{leetcode_base_url}/bitwise-and-of-numbers-range/[201. 数字范围按位与^]
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|{doc_base_url}/0201-bitwise-and-of-numbers-range.adoc[题解]
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|⭕️ 位运算！暴力解法超时！可以利用 stem:[n&(n-1)] 去掉 `n` 的最后一位 `1`，反复计算，直到范围的左边界为止即可得到答案。
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|===
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截止目前，本轮练习一共完成 {codes2503} 道题。

‎src/main/java/com/diguage/algo/leetcode/_0201_BitwiseAndOfNumbersRange.java

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package com.diguage.algo.leetcode;
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public class _0201_BitwiseAndOfNumbersRange {
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// tag::answer[]
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/**
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* @author D瓜哥 · https://www.diguage.com
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* @since 2025-06-11 23:15:14
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*/
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public int rangeBitwiseAnd(int left, int right) {
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while (left < right) {
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right = right & (right - 1);
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}
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return right;
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}
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// end::answer[]
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}

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