diff --git a/notes/面试总结.md b/notes/面试总结.md
index 35b7aa3a..184a80d6 100644
--- a/notes/面试总结.md
+++ b/notes/面试总结.md
@@ -26,6 +26,8 @@
 * [21. 调整数组顺序使奇数位于偶数前面](#21-调整数组顺序使奇数位于偶数前面)
 * [22. 链表中倒数第 K 个结点](#22-链表中倒数第-k-个结点)
 * [23. 链表中环的入口结点](#23-链表中环的入口结点)
+* [24. 反转链表](#24-反转链表)
+* [25. 合并两个排序的链表](#25-合并两个排序的链表)
 
 * [参考文献](#参考文献)
 <!-- GFM-TOC -->
@@ -1790,6 +1792,179 @@ class Solution:
             if pHead in plist:
                 return pHead
 ```
+# 24. 反转链表
+
+[NowCoder](https://www.nowcoder.com/practice/75e878df47f24fdc9dc3e400ec6058ca?tpId=13&tqId=11168&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
+
+## 解题思路
+
+### 递归
+
+```java
+public ListNode ReverseList(ListNode head) {
+    if (head == null || head.next == null)
+        return head;
+    ListNode next = head.next;
+    head.next = null;
+    ListNode newHead = ReverseList(next);
+    next.next = head;
+    return newHead;
+}
+```
+
+```python
+# -*- coding:utf-8 -*-
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+class Solution:
+    # 返回ListNode
+    def ReverseList(self, pHead):
+        # write code here
+        if pHead == None or pHead.next == None:
+            return pHead
+        next = pHead.next
+        pHead.next = None
+        newHead = self.ReverseList(next)
+        next.next = pHead
+        return newHead
+```
+
+### 迭代
+
+```java
+public ListNode ReverseList(ListNode head) {
+    ListNode newList = new ListNode(-1);
+    while (head != null) {
+        ListNode next = head.next;
+        head.next = newList.next;
+        newList.next = head;
+        head = next;
+    }
+    return newList.next;
+}
+```
+
+```python
+# -*- coding:utf-8 -*-
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+class Solution:
+    # 返回ListNode
+    def ReverseList(self, pHead):
+        # write code here
+        newList = ListNode(-1)
+        while pHead:
+            next = pHead.next
+            pHead.next = newList.next
+            newList.next = pHead
+            pHead = next
+        return newList.next
+```
+
+# 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)
+
+## 题目描述
+
+<div align="center"> <img src="../pics//43f2cafa-3568-4a89-a895-4725666b94a6.png" width="500"/> </div><br>
+
+## 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;
+    }
+}
+```
+
+```python
+# -*- coding:utf-8 -*-
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+class Solution:
+    # 返回合并后列表
+    def Merge(self, pHead1, pHead2):
+        # write code here
+        if pHead1 == None:
+            return pHead2
+        if pHead2 == None:
+            return pHead1
+        if pHead1.val <= pHead2.val:
+            pHead1.next = self.Merge(pHead1.next, pHead2)
+            return pHead1
+        else:
+            pHead2.next = self.Merge(pHead1, pHead2.next)
+            return pHead2
+```
+
+### 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;
+}
+```
+
+```python
+# -*- coding:utf-8 -*-
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+class Solution:
+    # 返回合并后列表
+    def Merge(self, pHead1, pHead2):
+        # write code here
+        head = ListNode(-1)
+        cur = head
+        while pHead1 and pHead2:
+            if pHead1.val <= pHead2.val:
+                cur.next = pHead1
+                pHead1 = pHead1.next
+            else:
+                cur.next = pHead2
+                pHead2 = pHead2.next
+            cur = cur.next
+        if pHead1:
+            cur.next = pHead1
+        if pHead2:
+            cur.next = pHead2
+        return head.next
+```
 
 # 参考文献