Merge pull request #688 from bourne-3/bourneLee

0234-回文链表Java实现

problems/0234.回文链表.md

@@ -144,6 +144,75 @@ public:
 ## Java
 
 ```java
+// 方法一，使用数组
+class Solution {
+    public boolean isPalindrome(ListNode head) {
+        int len = 0;
+        // 统计链表长度
+        ListNode cur = head;
+        while (cur != null) {
+            len++;
+            cur = cur.next;
+        }
+        cur = head;
+        int[] res = new int[len];
+        // 将元素加到数组之中
+        for (int i = 0; i < res.length; i++){
+            res[i] = cur.val;
+            cur = cur.next;
+        }
+        // 比较回文
+        for (int i = 0, j = len - 1; i < j; i++, j--){
+            if (res[i] != res[j]){
+                return false;
+            }
+        }
+        return true;
+    }
+}
+
+// 方法二，快慢指针
+class Solution {
+    public boolean isPalindrome(ListNode head) {
+        // 如果为空或者仅有一个节点，返回true
+        if (head == null && head.next == null) return true;
+        ListNode slow = head;
+        ListNode fast = head;
+        ListNode pre = head;
+        while (fast != null && fast.next != null){
+            pre = slow;  // 记录slow的前一个结点
+            slow = slow.next;
+            fast = fast.next.next;
+        }
+        pre.next = null;  // 分割两个链表
+
+        // 前半部分
+        ListNode cur1 = head;
+        // 后半部分。这里使用了反转链表
+        ListNode cur2 = reverseList(slow);
+
+        while (cur1 != null){
+            if (cur1.val != cur2.val) return false;
+
+            // 注意要移动两个结点
+            cur1 = cur1.next;
+            cur2 = cur2.next;
+        }
+        return true;
+    }
+    ListNode reverseList(ListNode head){
+        // 反转链表
+        ListNode tmp = null;
+        ListNode pre = null;
+        while (head != null){
+            tmp = head.next;
+            head.next = pre;
+            pre = head;
+            head = tmp;
+        }
+        return pre;
+    }
+}
 ```
 
 ## Python
@@ -209,11 +278,13 @@ class Solution:
 ## Go
 
 ```go
+
 ```
 
 ## JavaScript
 
 ```js
+
 ```