三刷139

Author: diguage

docs/0139-word-break.adoc

@@ -1,7 +1,7 @@
 [#0139-word-break]
 = 139. 单词拆分
 
-https://leetcode.cn/problems/word-break/[LeetCode - 139. 单词拆分 ^]
+https://leetcode.cn/problems/word-break/[LeetCode - 139. 单词拆分^]
 
 给你一个字符串 `s` 和一个字符串列表 `wordDict` 作为字典。如果可以利用字典中出现的一个或多个单词拼接出 `s` 则返回 `true`。
 
@@ -44,22 +44,24 @@ https://leetcode.cn/problems/word-break/[LeetCode - 139. 单词拆分 ^]
 
 回溯+备忘录：
 
-image::images/0139-01.png[{image_attr}]
+image::images/0139-10.png[{image_attr}]
 
 动态规划
 
-image::images/0139-02.png[{image_attr}]
+image::images/0139-11.png[{image_attr}]
 
-image::images/0139-03.png[{image_attr}]
+image::images/0139-12.png[{image_attr}]
 
-image::images/0139-04.png[{image_attr}]
+image::images/0139-13.png[{image_attr}]
 
 我的思路：
 
-stem:[dp\[i+j\] = dp\[i\] & s\[i, i+j\] in dict]
+stem:[dp[i+j\] = dp[i\] \& (s[i, i+j\] \in  dict)]
 
 从 stem:[0] 到 stem:[i] 已经确定。从当前位置 stem:[i] 向前推进到 stem:[i+j]，其中 stem:[j] 是某个一个单词的长度。
 
+动态规划。有两种思路：①向后看，从当前位置截取 `word` 长度的子串，判断子串和当前 `word` 是否相等以及之前的字符串是否可以可以拆分；②向前进，从零开始，根据当前 `word` 的长度，在同等长度子串相等的情况下，向前设置子串可以拆分，直到结尾。
+
 [[src-0139]]
 [tabs]
 ====
@@ -89,9 +91,20 @@ include::{sourcedir}/_0139_WordBreak_20.java[tag=answer]
 include::{sourcedir}/_0139_WordBreak_21.java[tag=answer]
 ----
 --
+
+三刷::
++
+--
+[{java_src_attr}]
+----
+include::{sourcedir}/_0139_WordBreak_3.java[tag=answer]
+----
+--
 ====
 
 == 参考资料
 
+. https://leetcode.cn/problems/word-break/solutions/2968135/jiao-ni-yi-bu-bu-si-kao-dpcong-ji-yi-hua-chrs/[139. 单词拆分 - 教你一步步思考 DP：从记忆化搜索到递推，附题单！^]
 . https://leetcode.cn/problems/word-break/solutions/302779/shou-hui-tu-jie-san-chong-fang-fa-dfs-bfs-dong-tai/[139. 单词拆分 - 「手画图解」剖析三种解法: DFS, BFS, 动态规划^] -- 这个题解非常好，从回溯，到加备忘录
-. https://leetcode.cn/problems/word-break/solutions/50986/dong-tai-gui-hua-ji-yi-hua-hui-su-zhu-xing-jie-shi/[139. 单词拆分 - 动态规划+记忆化回溯 逐行解释 Python3^]
+. https://leetcode.cn/problems/word-break/solutions/50986/dong-tai-gui-hua-ji-yi-hua-hui-su-zhu-xing-jie-shi/[139. 单词拆分 - 动态规划+记忆化回溯 逐行解释^]
+. https://leetcode.cn/problems/word-break/solutions/850456/dai-ma-sui-xiang-lu-139-dan-ci-chai-fen-50a1a/[139. 单词拆分 - 带你学透完全背包！^]

docs/0141-linked-list-cycle.adoc

@@ -1,7 +1,7 @@
 [#0141-linked-list-cycle]
 = 141. 环形链表
 
-https://leetcode.cn/problems/linked-list-cycle/[LeetCode - 141. 环形链表 ^]
+https://leetcode.cn/problems/linked-list-cycle/[LeetCode - 141. 环形链表^]
 
 给你一个链表的头节点 `head` ，判断链表中是否有环。
 
@@ -97,8 +97,9 @@ include::{sourcedir}/_0141_LinkedListCycle_3.java[tag=answer]
 
 == 参考资料
 
-. https://leetcode.cn/problems/linked-list-cycle/solutions/440042/huan-xing-lian-biao-by-leetcode-solution/[141. 环形链表 - 官方题解^]
+. https://leetcode.cn/problems/linked-list-cycle/solutions/1999269/mei-xiang-ming-bai-yi-ge-shi-pin-jiang-t-c4sw/[141. 环形链表 - 没想明白？一个视频讲透快慢指针！^]
 . https://leetcode.cn/problems/linked-list-cycle/solutions/175734/yi-wen-gao-ding-chang-jian-de-lian-biao-wen-ti-h-2/[141. 环形链表 - 一文搞定常见的链表问题^]
-
+. https://leetcode.cn/problems/linked-list-cycle/solutions/440042/huan-xing-lian-biao-by-leetcode-solution/[141. 环形链表 - 官方题解^]
+. https://leetcode.cn/problems/linked-list-cycle/solutions/438358/3chong-jie-jue-fang-shi-liang-chong-ji-bai-liao-10/[141. 环形链表 - 9种解决方式，8种击败了100%的用户^]
 
 

docs/0142-linked-list-cycle-ii.adoc

@@ -139,8 +139,7 @@ include::{sourcedir}/_0142_LinkedListCycleII_4.java[tag=answer]
 
 == 参考资料
 
-. https://en.wikipedia.org/wiki/Cycle_detection[Cycle detection - Wikipedia^]
-. https://leetcode.cn/problems/linked-list-cycle-ii/solutions/441131/huan-xing-lian-biao-ii-by-leetcode-solution/[142. 环形链表 II - 官方题解^]
+. https://leetcode.cn/problems/linked-list-cycle-ii/solutions/1999271/mei-xiang-ming-bai-yi-ge-shi-pin-jiang-t-nvsq/[142. 环形链表 II - 【图解】一张图秒懂环形链表 II，Floyd 判圈算法^]
 . https://leetcode.cn/problems/linked-list-cycle-ii/solutions/12616/linked-list-cycle-ii-kuai-man-zhi-zhen-shuang-zhi-/[142. 环形链表 II - 双指针，清晰图解^]
-
-
+. https://leetcode.cn/problems/linked-list-cycle-ii/solutions/441131/huan-xing-lian-biao-ii-by-leetcode-solution/[142. 环形链表 II - 官方题解^]
+. https://en.wikipedia.org/wiki/Cycle_detection[Cycle detection - Wikipedia^]

docs/images/0139-01.png renamed to docs/images/0139-10.png

@@ -1844,6 +1844,11 @@ endif::[]
 |{doc_base_url}/0146-lru-cache.adoc[题解]
 |✅ 链表+哈希。
 
+|{counter:codes2503}
+|{leetcode_base_url}/word-break/[139. 单词拆分^]
+|{doc_base_url}/0139-word-break.adoc[题解]
+|✅ 动态规划。有两种思路：①向后看，从当前位置截取 `word` 长度的子串，判断子串和当前 `word` 是否相等以及之前的字符串是否可以可以拆分；②向前进，从零开始，根据当前 `word` 的长度，在同等长度子串相等的情况下，向前设置子串可以拆分，直到结尾。
+
 |===
 
 截止目前，本轮练习一共完成 {codes2503} 道题。

docs/images/0139-02.png renamed to docs/images/0139-11.png

@@ -0,0 +1,35 @@
+package com.diguage.algo.leetcode;
+
+import java.util.List;
+
+public class _0139_WordBreak_3 {
+  // tag::answer[]
+
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2025-11-05 21:17:21
+   */
+  public boolean wordBreak(String s, List<String> wordDict) {
+    boolean[] dp = new boolean[s.length() + 1];
+    dp[0] = true;
+    for (int i = 1; i <= s.length(); i++) {
+      for (String word : wordDict) {
+        if (i < word.length()) {
+          continue;
+        }
+        dp[i] = dp[i - word.length()]
+          && word.equals(s.substring(i - word.length(), i));
+        if (dp[i]) {
+          break;
+        }
+      }
+    }
+    return dp[s.length()];
+  }
+  // end::answer[]
+
+  static void main() {
+    new _0139_WordBreak_3()
+      .wordBreak("leetcode", List.of("leet", "code"));
+  }
+}