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docs/notes/25. 合并两个排序的链表.md

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+# 25. 合并两个排序的链表
+
+[NowCoder](https://www.nowcoder.com/practice/d8b6b4358f774294a89de2a6ac4d9337?tpId=13&tqId=11169&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
+
+## 题目描述
+
+<img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/c094d2bc-ec75-444b-af77-d369dfb6b3b4.png" width="400"/>
+
+## 解题思路
+
+### 递归
+
+```java
+public ListNode Merge(ListNode list1, ListNode list2) {
+    if (list1 == null)
+        return list2;
+    if (list2 == null)
+        return list1;
+    if (list1.val <= list2.val) {
+        list1.next = Merge(list1.next, list2);
+        return list1;
+    } else {
+        list2.next = Merge(list1, list2.next);
+        return list2;
+    }
+}
+```
+
+### 迭代
+
+```java
+public ListNode Merge(ListNode list1, ListNode list2) {
+    ListNode head = new ListNode(-1);
+    ListNode cur = head;
+    while (list1 != null && list2 != null) {
+        if (list1.val <= list2.val) {
+            cur.next = list1;
+            list1 = list1.next;
+        } else {
+            cur.next = list2;
+            list2 = list2.next;
+        }
+        cur = cur.next;
+    }
+    if (list1 != null)
+        cur.next = list1;
+    if (list2 != null)
+        cur.next = list2;
+    return head.next;
+}
+```