四刷21

Author: diguage

Diff for: docs/0021-merge-two-sorted-lists.adoc

@@ -1,18 +1,38 @@
 [#0021-merge-two-sorted-lists]
-= 21. Merge Two Sorted Lists
+= 21. 合并两个有序链表
 
-{leetcode}/problems/merge-two-sorted-lists/[LeetCode - Merge Two Sorted Lists^]
+https://leetcode.cn/problems/merge-two-sorted-lists/[LeetCode - 21. 合并两个有序链表 ^]
 
-Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.
+将两个升序链表合并为一个新的 *升序* 链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。
+
+*示例 1：*
 
 image::images/0021-00.jpg[{image_attr}]
 
-*Example:*
-[subs="verbatim,quotes,macros"]
-----
-*Input:* 1->2->4, 1->3->4
-*Output:* 1->1->2->3->4->4
-----
+....
+输入：l1 = [1,2,4], l2 = [1,3,4]
+输出：[1,1,2,3,4,4]
+....
+
+*示例 2：*
+
+....
+输入：l1 = [], l2 = []
+输出：[]
+....
+
+*示例 3：*
+
+....
+输入：l1 = [], l2 = [0]
+输出：[0]
+....
+
+*提示：*
+
+* 两个链表的节点数目范围是 `[0, 50]`
+* `+-100 <= Node.val <= 100+`
+* `l1` 和 `l2` 均按 *非递减顺序* 排列
 
 == 思路分析
 
@@ -69,6 +89,15 @@ include::{sourcedir}/_0021_MergeTwoSortedLists_2.java[tag=answer]
 include::{sourcedir}/_0021_MergeTwoSortedLists_3.java[tag=answer]
 ----
 --
+
+四刷::
++
+--
+[{java_src_attr}]
+----
+include::{sourcedir}/_0021_MergeTwoSortedLists_4.java[tag=answer]
+----
+--
 ====
 
 

Diff for: logbook/202503.adoc

@@ -150,6 +150,11 @@ endif::[]
 |{doc_base_url}/0053-maximum-subarray.adoc[题解]
 |⭕️ 动态规划。稀里糊涂就对了，还要对推演。另有更精妙的分治解法，抽空再尝试。
 
+|{counter:codes2503}
+|{leetcode_base_url}/merge-two-sorted-lists/[21. Merge Two Sorted Lists^]
+|{doc_base_url}/0021-merge-two-sorted-lists.adoc[题解]
+|✅
+
 |===
 
 截止目前，本轮练习一共完成 {codes2503} 道题。

Diff for: src/main/java/com/diguage/algo/leetcode/_0021_MergeTwoSortedLists_4.java

@@ -0,0 +1,33 @@
+package com.diguage.algo.leetcode;
+
+import com.diguage.algo.util.ListNode;
+
+public class _0021_MergeTwoSortedLists_4 {
+  // tag::answer[]
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2025-04-05 23:39:31
+   */
+  public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
+    ListNode dummy = new ListNode(0);
+    ListNode cur = dummy;
+    while (l1 != null && l2 != null) {
+      if (l1.val <= l2.val) {
+        cur.next = l1;
+        l1 = l1.next;
+      } else {
+        cur.next = l2;
+        l2 = l2.next;
+      }
+      cur = cur.next;
+    }
+    if (l1 != null) {
+      cur.next = l1;
+    }
+    if (l2 != null) {
+      cur.next = l2;
+    }
+    return dummy.next;
+  }
+  // end::answer[]
+}