三刷21

Author: diguage

docs/0021-merge-two-sorted-lists.adoc

@@ -16,22 +16,9 @@ image::images/0021-00.jpg[{image_attr}]
 
 == 思路分析
 
-=== 迭代
+最先想到的是迭代。
 
-[[src-0021]]
-[{java_src_attr}]
-----
-include::{sourcedir}/_0021_MergeTwoSortedLists.java[tag=answer]
-----
-
-[{java_src_attr}]
-----
-include::{sourcedir}/_0021_MergeTwoSortedLists_2.java[tag=answer]
-----
-
-=== 递归
-
-没想到这个题也可以用递归解决！
+但是，也可以用递归解决！
 
 image::images/0021-09.png[{image_attr}]
 
@@ -53,6 +40,37 @@ image::images/0021-07.jpeg[{image_attr}]
 
 image::images/0021-08.jpeg[{image_attr}]
 
+[[src-0021]]
+[tabs]
+====
+一刷::
++
+--
+[{java_src_attr}]
+----
+include::{sourcedir}/_0021_MergeTwoSortedLists.java[tag=answer]
+----
+--
+
+二刷::
++
+--
+[{java_src_attr}]
+----
+include::{sourcedir}/_0021_MergeTwoSortedLists_2.java[tag=answer]
+----
+--
+
+三刷::
++
+--
+[{java_src_attr}]
+----
+include::{sourcedir}/_0021_MergeTwoSortedLists_3.java[tag=answer]
+----
+--
+====
+
 
 == 参考资料
 

logbook/202406.adoc

@@ -600,6 +600,11 @@
 |{doc_base_url}/0343-integer-break.adoc[题解]
 |动态规划，数学分析也可以看一下：怎么样计算自然底数 `e` 的值？
 
+|{counter:codes}
+|{leetcode_base_url}/merge-two-sorted-lists/[21. Merge Two Sorted Lists^]
+|{doc_base_url}/0021-merge-two-sorted-lists.adoc[题解]
+|递归
+
 |===
 
 截止目前，本轮练习一共完成 {codes} 道题。

src/main/java/com/diguage/algo/leetcode/_0021_MergeTwoSortedLists_2.java

@@ -27,6 +27,10 @@
  */
 public class _0021_MergeTwoSortedLists_2 {
   // tag::answer[]
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2024-07-02 19:43:59
+   */
   public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
     if (Objects.isNull(l1)) {
       return l2;

src/main/java/com/diguage/algo/leetcode/_0021_MergeTwoSortedLists_3.java

@@ -0,0 +1,26 @@
+package com.diguage.algo.leetcode;
+
+import com.diguage.algo.util.ListNode;
+
+public class _0021_MergeTwoSortedLists_3 {
+  // tag::answer[]
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2024-07-02 19:43:59
+   */
+  public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
+    if (l1 == null) {
+      return l2;
+    } else if (l2 == null) {
+      return l1;
+    } else if (l1.val < l2.val) {
+      // 升序链表，则把最小的留下，其余的进行递归。
+      l1.next = mergeTwoLists(l1.next, l2);
+      return l1;
+    } else {
+      l2.next = mergeTwoLists(l1, l2.next);
+      return l2;
+    }
+  }
+  // end::answer[]
+}