一刷822

Author: diguage

README.adoc

@@ -5779,13 +5779,13 @@ TIP: **公众号的微信号是: `jikerizhi`**。__因为众所周知的原因
 |Easy
 |
 
-//|{counter:codes}
-//|{leetcode_base_url}/card-flipping-game/[822. Card Flipping Game^]
-//|{source_base_url}/_0822_CardFlippingGame.java[Java]
-//|{doc_base_url}/0822-card-flipping-game.adoc[题解]
-//|Medium
-//|
-//
+|{counter:codes}
+|{leetcode_base_url}/card-flipping-game/[822. Card Flipping Game^]
+|{source_base_url}/_0822_CardFlippingGame.java[Java]
+|{doc_base_url}/0822-card-flipping-game.adoc[题解]
+|Medium
+|
+
 //|{counter:codes}
 //|{leetcode_base_url}/binary-trees-with-factors/[823. Binary Trees With Factors^]
 //|{source_base_url}/_0823_BinaryTreesWithFactors.java[Java]

docs/0822-card-flipping-game.adoc

@@ -1,45 +1,74 @@
 [#0822-card-flipping-game]
-= 822. Card Flipping Game
+= 822. 翻转卡片游戏
 
-{leetcode}/problems/card-flipping-game/[LeetCode - Card Flipping Game^]
+https://leetcode.cn/problems/card-flipping-game/[LeetCode - 822. 翻转卡片游戏^]
 
-On a table are `N` cards, with a positive integer printed on the front and back of each card (possibly different).
+在桌子上有 `n` 张卡片，每张卡片的正面和背面都写着一个正数（正面与背面上的数有可能不一样）。
 
-We flip any number of cards, and after we choose one card. 
+我们可以先翻转任意张卡片，然后选择其中一张卡片。
 
-If the number `X` on the back of the chosen card is not on the front of any card, then this number X is good.
+如果选中的那张卡片背面的数字 `x` 与任意一张卡片的正面的数字都不同，那么这个数字是我们想要的数字。
 
-What is the smallest number that is good?  If no number is good, output `0`.
+哪个数是这些想要的数字中最小的数（找到这些数中的最小值）呢？如果没有一个数字符合要求的，输出 `0`。
 
-Here, `fronts[i]` and `backs[i]` represent the number on the front and back of card `i`. 
+其中, `fronts[i]` 和 `backs[i]` 分别代表第 `i` 张卡片的正面和背面的数字。
 
-A flip swaps the front and back numbers, so the value on the front is now on the back and vice versa.
+如果我们通过翻转卡片来交换正面与背面上的数，那么当初在正面的数就变成背面的数，背面的数就变成正面的数。
 
-*Example:*
+*示例 1：*
 
-[subs="verbatim,quotes,macros"]
-----
-*Input:* fronts = [1,2,4,4,7], backs = [1,3,4,1,3]
-*Output:* `2`
-*Explanation:* If we flip the second card, the fronts are `[1,3,4,4,7]` and the backs are `[1,2,4,1,3]`.
-We choose the second card, which has number 2 on the back, and it isn't on the front of any card, so `2` is good.
-----
+....
+输入：fronts = [1,2,4,4,7], backs = [1,3,4,1,3]
+输出：2
+解释：假设我们翻转第二张卡片，那么在正面的数变成了 [1,3,4,4,7] ， 背面的数变成了 [1,2,4,1,3]。
+接着我们选择第二张卡片，因为现在该卡片的背面的数是 2，2 与任意卡片上正面的数都不同，所以 2 就是我们想要的数字。
+....
+
+*示例 2：*
 
- 
+....
+输入：fronts = [1], backs = [1]
+输出：0
+解释：
+无论如何翻转都无法得到想要的数字，所以返回 0 。
+....
 
-*Note:*
+*提示：*
 
+* `n == fronts.length == backs.length`
+* `1 \<= n \<= 1000`
+* `1 \<= fronts[i], backs[i] \<= 2000`
 
-. `1 <= fronts.length == backs.length <= 1000`.
-. `1 <= fronts[i] <= 2000`.
-. `1 <= backs[i] <= 2000`.
 
 
+== 思路分析
 
+首先，排除的是正反面一样的数字。剩下的数字里，找最小的数字即可。有点想复杂了。
 
 [[src-0822]]
+[tabs]
+====
+一刷::
++
+--
 [{java_src_attr}]
 ----
 include::{sourcedir}/_0822_CardFlippingGame.java[tag=answer]
 ----
+--
+
+// 二刷::
+// +
+// --
+// [{java_src_attr}]
+// ----
+// include::{sourcedir}/_0822_CardFlippingGame_2.java[tag=answer]
+// ----
+// --
+====
+
+
+== 参考资料
 
+. https://leetcode.cn/problems/card-flipping-game/solutions/2368863/yue-du-li-jie-ti-pythonjavacgojs-by-endl-ze7f/[822. 翻转卡片游戏 - 阅读理解题^]
+. https://leetcode.cn/problems/card-flipping-game/solutions/2365854/fan-zhuan-qia-pian-you-xi-by-leetcode-so-acbj/[822. 翻转卡片游戏 - 官方题解^]

docs/index.adoc

@@ -1736,7 +1736,7 @@ include::0820-short-encoding-of-words.adoc[leveloffset=+1]
 
 include::0821-shortest-distance-to-a-character.adoc[leveloffset=+1]
 
-// include::0822-card-flipping-game.adoc[leveloffset=+1]
+include::0822-card-flipping-game.adoc[leveloffset=+1]
 
 // include::0823-binary-trees-with-factors.adoc[leveloffset=+1]
 

logbook/202601.adoc

@@ -1021,6 +1021,12 @@ endif::[]
 |{doc_base_url}/0820-short-encoding-of-words.adoc[题解]
 |✅ 排序。按长度倒序，后根据判断是否存在在决定是否插入。更优解是前缀树。
 
+|{counter:codes}
+|{leetcode_base_url}/card-flipping-game/[822. 翻转卡片游戏^]
+|{doc_base_url}/0822-card-flipping-game.adoc[题解]
+|❌ 哈希。想复杂了。排除的是正反面一样的数字。剩下的数字里，找最小的数字即可。
+
+
 
 |===
 

logbook/corrections.adoc

@@ -1233,6 +1233,11 @@ endif::[]
 |{doc_base_url}/0821-shortest-distance-to-a-character.adoc[题解]
 |❌
 
+|{counter:codes}
+|{leetcode_base_url}/card-flipping-game/[822. 翻转卡片游戏^]
+|{doc_base_url}/0822-card-flipping-game.adoc[题解]
+|❌
+
 |===
 
 截止目前，本轮练习一共完成 {corrections} 道题。

src/main/java/com/diguage/algo/leetcode/_0822_CardFlippingGame.java

@@ -0,0 +1,34 @@
+package com.diguage.algo.leetcode;
+
+import java.util.HashSet;
+import java.util.Set;
+
+public class _0822_CardFlippingGame {
+  // tag::answer[]
+
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2026-07-15 22:52:49
+   */
+  public int flipgame(int[] fronts, int[] backs) {
+    Set<Integer> set = new HashSet<>();
+    for (int i = 0; i < fronts.length; i++) {
+      if (fronts[i] == backs[i]) {
+        set.add(fronts[i]);
+      }
+    }
+    int result = Integer.MAX_VALUE;
+    for (int front : fronts) {
+      if (!set.contains(front)) {
+        result = Math.min(result, front);
+      }
+    }
+    for (int back : backs) {
+      if (!set.contains(back)) {
+        result = Math.min(result, back);
+      }
+    }
+    return result == Integer.MAX_VALUE ? 0 : result;
+  }
+  // end::answer[]
+}