一刷231

Author: diguage

README.adoc

@@ -1643,13 +1643,13 @@ TIP: **公众号的微信号是: `jikerizhi`**。__因为众所周知的原因
 |Medium
 |
 
-//|{counter:codes}
-//|{leetcode_base_url}/power-of-two/[231. Power of Two^]
-//|{source_base_url}/_0231_PowerOfTwo.java[Java]
-//|{doc_base_url}/0231-power-of-two.adoc[题解]
-//|Easy
-//|
-//
+|{counter:codes}
+|{leetcode_base_url}/power-of-two/[231. Power of Two^]
+|{source_base_url}/_0231_PowerOfTwo.java[Java]
+|{doc_base_url}/0231-power-of-two.adoc[题解]
+|Easy
+|
+
 //|{counter:codes}
 //|{leetcode_base_url}/implement-queue-using-stacks/[232. Implement Queue using Stacks^]
 //|{source_base_url}/_0232_ImplementQueueUsingStacks.java[Java]

docs/0231-power-of-two.adoc

@@ -32,11 +32,38 @@ Given an integer, write a function to determine if it is a power of two.
 *Output:* false
 ----
 
+== 思路分析
 
+这道题可以利用 xref:0191-number-of-1-bits.adoc[191. Number of 1 Bits] 中的思路。
+
+2 的幂次方数，它们的二进制数只有一个 `1`，那么 `n & (n - 1)` 必然为 `0`。
+
+看答案，也可以使用约数法。
 
 [[src-0231]]
+[tabs]
+====
+一刷::
++
+--
 [{java_src_attr}]
 ----
 include::{sourcedir}/_0231_PowerOfTwo.java[tag=answer]
 ----
+--
+
+// 二刷::
+// +
+// --
+// [{java_src_attr}]
+// ----
+// include::{sourcedir}/_0231_PowerOfTwo_2.java[tag=answer]
+// ----
+// --
+====
+
+== 参考资料
+
+. https://leetcode.cn/problems/power-of-two/solutions/796201/2de-mi-by-leetcode-solution-rny3/?envType=study-plan-v2&envId=selected-coding-interview[231. 2 的幂 - 官方题解^]
+. https://leetcode.cn/problems/number-of-1-bits/solutions/2361978/191-wei-1-de-ge-shu-wei-yun-suan-qing-xi-40rw/?envType=study-plan-v2&envId=selected-coding-interview[191. 位1的个数 - 位运算，清晰图解^]
 

docs/index.adoc

@@ -536,7 +536,7 @@ include::0227-basic-calculator-ii.adoc[leveloffset=+1]
 
 include::0230-kth-smallest-element-in-a-bst.adoc[leveloffset=+1]
 
-// include::0231-power-of-two.adoc[leveloffset=+1]
+include::0231-power-of-two.adoc[leveloffset=+1]
 
 // include::0232-implement-queue-using-stacks.adoc[leveloffset=+1]
 

logbook/202406.adoc

@@ -590,6 +590,12 @@
 |{doc_base_url}/0240-search-a-2d-matrix-ii.adoc[题解]
 |不能对矩阵做二分查找，可以对单行做二分查找。要分析题目，查看如何排除不符合要求的元素。
 
+|{counter:codes}
+|{leetcode_base_url}/power-of-two/[231. Power of Two^]
+|{doc_base_url}/0231-power-of-two.adoc[题解]
+|位运算， `n & (n - 1)` 或约数
+
+
 |===
 
 截止目前，本轮练习一共完成 {codes} 道题。

src/main/java/com/diguage/algo/leetcode/_0231_PowerOfTwo.java

@@ -0,0 +1,13 @@
+package com.diguage.algo.leetcode;
+
+public class _0231_PowerOfTwo {
+  // tag::answer[]
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2024-09-15 20:19:55
+   */
+  public boolean isPowerOfTwo(int n) {
+    return n > 0 && (n & (n - 1)) == 0;
+  }
+  // end::answer[]
+}