添加cpp，py和js （上半部分）

animation-simulation/链表篇/234. 回文链表.md

@@ -35,6 +35,8 @@
 
 **题目代码**
 
+Java Code:
+
 ```java
 class Solution {
     public boolean isPalindrome(ListNode head) {
@@ -64,6 +66,83 @@ class Solution {
 
 ```
 
+C++ Code:
+
+```cpp
+class Solution {
+public:
+    bool isPalindrome(ListNode* head) {
+        vector<int> arr;
+        ListNode* copynode = head;
+        while (copynode) {
+            arr.push_back(copynode->val);
+            copynode = copynode->next;
+        }
+        int back = 0;
+        int pro = arr.size() - 1;
+        while (back < pro) {
+            if (arr[back] != arr[pro]) {
+                return false;
+            }
+            back++;
+            pro--;
+        }
+        return true;
+    }
+};
+```
+
+JS Code:
+
+```js
+var isPalindrome = function(head) {
+    let arr = [];
+    let copynode = head;
+    //将链表的值复制到数组中
+    while (copynode) {
+        arr.push(copynode.val)
+        copynode = copynode.next
+    }
+    //双指针遍历数组
+    let back = 0;
+    let pro = arr.length - 1;
+    while (back < pro) {
+        //判断两个指针的值是否相等
+        if (arr[back] !== arr[pro]) {
+            return false
+        }
+        //移动指针
+        back += 1
+        pro -= 1
+    }
+    return true
+};
+```
+
+Python Code:
+
+```py
+class Solution:
+    def isPalindrome(self, head: ListNode) -> bool:
+        arr = []
+        copynode = head
+        # 将链表的值复制到数组中
+        while copynode is not None:
+            arr.append(copynode.val)
+            copynode = copynode.next
+        # 双指针遍历数组
+        back = 0
+        pro = len(arr) - 1
+        while back < pro:
+        	# 判断两个指针的值是否相等
+            if arr[back] != arr[pro]:
+                return False
+            # 移动指针
+            back += 1
+            pro -= 1
+        return True
+```
+
 这个方法可以直接通过，但是这个方法需要辅助数组，那我们还有其他更好的方法吗？
 
 双指针翻转链表法