diff --git a/notes/25. 合并两个排序的链表.md b/notes/25. 合并两个排序的链表.md
index ba0e6330..93472516 100644
--- a/notes/25. 合并两个排序的链表.md	
+++ b/notes/25. 合并两个排序的链表.md	
@@ -1,6 +1,6 @@
 # 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)
+[Leetcode](https://leetcode.cn/problems/he-bing-liang-ge-pai-xu-de-lian-biao-lcof/)
 
 ## 题目描述
 
@@ -11,17 +11,17 @@
 ### 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;
+public ListNode trainningPlan(ListNode l1, ListNode l2) {
+    if (l1 == null)
+        return l2;
+    if (l2 == null)
+        return l1;
+    if (l1.val <= l2.val) {
+        l1.next = trainningPlan(l1.next, l2);
+        return l1;
     } else {
-        list2.next = Merge(list1, list2.next);
-        return list2;
+        l2.next = trainningPlan(l1, l2.next);
+        return l2;
     }
 }
 ```
@@ -29,23 +29,23 @@ public ListNode Merge(ListNode list1, ListNode list2) {
 ### 迭代
 
 ```java
-public ListNode Merge(ListNode list1, ListNode list2) {
+public ListNode trainningPlan(ListNode l1, ListNode l2) {
     ListNode head = new ListNode(-1);
     ListNode cur = head;
-    while (list1 != null && list2 != null) {
-        if (list1.val <= list2.val) {
-            cur.next = list1;
-            list1 = list1.next;
+    while (l1 != null && l2 != null) {
+        if (l1.val <= l2.val) {
+            cur.next = l1;
+            l1 = l1.next;
         } else {
-            cur.next = list2;
-            list2 = list2.next;
+            cur.next = l2;
+            l2 = l2.next;
         }
         cur = cur.next;
     }
-    if (list1 != null)
-        cur.next = list1;
-    if (list2 != null)
-        cur.next = list2;
+    if (l1 != null)
+        cur.next = l1;
+    if (l2 != null)
+        cur.next = l2;
     return head.next;
 }
 ```