一刷869

Author: diguage

README.adoc

@@ -6108,13 +6108,13 @@ TIP: **公众号的微信号是: `jikerizhi`**。__因为众所周知的原因
 //|{doc_base_url}/0868-binary-gap.adoc[题解]
 //|Easy
 //|
-//
-//|{counter:codes}
-//|{leetcode_base_url}/reordered-power-of-2/[869. Reordered Power of 2^]
-//|{source_base_url}/_0869_ReorderedPowerOf2.java[Java]
-//|{doc_base_url}/0869-reordered-power-of-2.adoc[题解]
-//|Medium
-//|
+
+|{counter:codes}
+|{leetcode_base_url}/reordered-power-of-2/[869. Reordered Power of 2^]
+|{source_base_url}/_0869_ReorderedPowerOf2.java[Java]
+|{doc_base_url}/0869-reordered-power-of-2.adoc[题解]
+|Medium
+|
 
 |{counter:codes}
 |{leetcode_base_url}/advantage-shuffle/[870. Advantage Shuffle^]

docs/0869-reordered-power-of-2.adoc

@@ -7,11 +7,6 @@ Starting with a positive integer `N`, we reorder the digits in any order (includ
 
 Return `true` if and only if we can do this in a way such that the resulting number is a power of 2.
 
- 
-
-
-
-
 
 *Example 1:*
 
@@ -30,7 +25,6 @@ Return `true` if and only if we can do this in a way such that the resulting num
 *Output:* false
 ----
 
-
 *Example 3:*
 
 [subs="verbatim,quotes,macros"]
@@ -39,7 +33,6 @@ Return `true` if and only if we can do this in a way such that the resulting num
 *Output:* true
 ----
 
-
 *Example 4:*
 
 [subs="verbatim,quotes,macros"]
@@ -48,7 +41,6 @@ Return `true` if and only if we can do this in a way such that the resulting num
 *Output:* false
 ----
 
-
 *Example 5:*
 
 [subs="verbatim,quotes,macros"]
@@ -57,24 +49,40 @@ Return `true` if and only if we can do this in a way such that the resulting num
 *Output:* true
 ----
 
- 
-
 *Note:*
 
-
-. `1 <= N <= 10^9`
-
-
-
-
+. `1 \<= N \<= 10^9`
 
 
+== 思路分析
 
+根据数字，对其对应的字符数组进行排序，然后再取得其对应的字符串。
 
+一次计算 2 的幂次方，然后在字符串长度相等的情况下，排序其对应的字符数组，再转化成字符串跟参数的字符串想比较。
 
 [[src-0869]]
+[tabs]
+====
+一刷::
++
+--
 [{java_src_attr}]
 ----
 include::{sourcedir}/_0869_ReorderedPowerOf2.java[tag=answer]
 ----
+--
+
+// 二刷::
+// +
+// --
+// [{java_src_attr}]
+// ----
+// include::{sourcedir}/_0869_ReorderedPowerOf2_2.java[tag=answer]
+// ----
+// --
+====
+
+== 参考资料
+
+. https://leetcode.cn/problems/reordered-power-of-2/solutions/1068761/zhong-xin-pai-xu-de-dao-2-de-mi-by-leetc-4fvs/[869. 重新排序得到 2 的幂 - 官方题解^]
 

docs/index.adoc

@@ -1812,7 +1812,7 @@ include::0796-rotate-string.adoc[leveloffset=+1]
 
 // include::0868-binary-gap.adoc[leveloffset=+1]
 
-// include::0869-reordered-power-of-2.adoc[leveloffset=+1]
+include::0869-reordered-power-of-2.adoc[leveloffset=+1]
 
 include::0870-advantage-shuffle.adoc[leveloffset=+1]
 

logbook/202406.adoc

@@ -875,6 +875,11 @@
 |⭕️ 贪心。排序规则跟 179、870 题类似：保存下标，使用其他数组的值对下标数组进行排序。
 
 
+|{counter:codes}
+|{leetcode_base_url}/reordered-power-of-2/[869. Reordered Power of 2^]
+|{doc_base_url}/0869-reordered-power-of-2.adoc[题解]
+|⭕️ 参考答案做了优化
+
 |===
 
 截止目前，本轮练习一共完成 {codes} 道题。

src/main/java/com/diguage/algo/leetcode/_0869_ReorderedPowerOf2.java

@@ -0,0 +1,32 @@
+package com.diguage.algo.leetcode;
+
+import java.util.Arrays;
+import java.util.Objects;
+
+public class _0869_ReorderedPowerOf2 {
+  // tag::answer[]
+
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2024-09-25 20:47:35
+   */
+  public boolean reorderedPowerOf2(int n) {
+    char[] chars = String.valueOf(n).toCharArray();
+    Arrays.sort(chars);
+    int pow = 1;
+    String str = new String(chars);
+    for (int i = 0; Math.pow(2, i - 1) < Integer.MAX_VALUE; i++) {
+      String ps = String.valueOf(pow);
+      if (ps.length() == chars.length) {
+        char[] pchars = ps.toCharArray();
+        Arrays.sort(pchars);
+        if (str.equals(new String(pchars))) {
+          return true;
+        }
+      }
+      pow *= 2;
+    }
+    return false;
+  }
+  // end::answer[]
+}