feat: add solutions to lcof2 problem: No.027.Palindrome Linked List

Author: yanglbme

lcof2/剑指 Offer II 026. 重排链表/README.md

@@ -51,8 +51,8 @@
 
 相当于这 3 道问题，只需要 5 行代码将它们组合：
 
-- [链表的中间结点](../../solution/0800-0899/0876.Middle%20of%20the%20Linked%20List/README.md)
-- [反转链表](../../solution/0200-0299/0206.Reverse%20Linked%20List/README.md)
+- [链表的中间结点](/solution/0800-0899/0876.Middle%20of%20the%20Linked%20List/README.md)
+- [反转链表](/solution/0200-0299/0206.Reverse%20Linked%20List/README.md)
 - 合并两个链表
 
 <!-- tabs:start -->
@@ -166,6 +166,71 @@ class Solution {
 }
 ```
 
+### **C++**
+
+```cpp
+/**
+ * Definition for singly-linked list.
+ * struct ListNode {
+ *     int val;
+ *     ListNode *next;
+ *     ListNode() : val(0), next(nullptr) {}
+ *     ListNode(int x) : val(x), next(nullptr) {}
+ *     ListNode(int x, ListNode *next) : val(x), next(next) {}
+ * };
+ */
+class Solution {
+public:
+    void reorderList(ListNode* head) {
+        ListNode* mid = middleNode(head);
+        ListNode* tmp = mid->next;
+        mid->next = nullptr;
+        tmp = reverseList(tmp);
+        head = mergeTwoLists(head, tmp);
+    }
+
+    ListNode* middleNode(ListNode* head) {
+        ListNode* slow = head;
+        ListNode* fast = head;
+        while (fast && fast->next)
+        {
+            slow = slow->next;
+            fast = fast->next->next;
+        }
+        return slow;
+    }
+
+    ListNode* reverseList(ListNode* head) {
+        ListNode* pre = nullptr;
+        ListNode* cur = head;
+        while (cur)
+        {
+            ListNode* tmp = cur->next;
+            cur->next = pre;
+            pre = cur;
+            cur = tmp;
+        }
+        return pre;
+    }
+
+    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
+        ListNode* dummy = new ListNode();
+        ListNode* cur = dummy;
+        while (l1 && l2)
+        {
+            cur->next = l1;
+            l1 = l1->next;
+            cur = cur->next;
+            cur->next = l2;
+            l2 = l2->next;
+            cur = cur->next;
+        }
+        cur->next = l1 ? l1 : l2;
+        return dummy->next;
+    }
+};
+```
+
 ### **Go**
 
 ```go

lcof2/剑指 Offer II 026. 重排链表/Solution.cpp

@@ -0,0 +1,60 @@
+/**
+ * Definition for singly-linked list.
+ * struct ListNode {
+ *     int val;
+ *     ListNode *next;
+ *     ListNode() : val(0), next(nullptr) {}
+ *     ListNode(int x) : val(x), next(nullptr) {}
+ *     ListNode(int x, ListNode *next) : val(x), next(next) {}
+ * };
+ */
+class Solution {
+public:
+    void reorderList(ListNode* head) {
+        ListNode* mid = middleNode(head);
+        ListNode* tmp = mid->next;
+        mid->next = nullptr;
+        tmp = reverseList(tmp);
+        head = mergeTwoLists(head, tmp);
+    }
+
+    ListNode* middleNode(ListNode* head) {
+        ListNode* slow = head;
+        ListNode* fast = head;
+        while (fast && fast->next)
+        {
+            slow = slow->next;
+            fast = fast->next->next;
+        }
+        return slow;
+    }
+
+    ListNode* reverseList(ListNode* head) {
+        ListNode* pre = nullptr;
+        ListNode* cur = head;
+        while (cur)
+        {
+            ListNode* tmp = cur->next;
+            cur->next = pre;
+            pre = cur;
+            cur = tmp;
+        }
+        return pre;
+    }
+
+    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
+        ListNode* dummy = new ListNode();
+        ListNode* cur = dummy;
+        while (l1 && l2)
+        {
+            cur->next = l1;
+            l1 = l1->next;
+            cur = cur->next;
+            cur->next = l2;
+            l2 = l2->next;
+            cur = cur->next;
+        }
+        cur->next = l1 ? l1 : l2;
+        return dummy->next;
+    }
+};