Add solution 234 [Java]

Author: yanglbme

README.md

@@ -36,6 +36,7 @@ Complete [solutions](https://github.com/doocs/leetcode/tree/master/solution) to
 | 198 | [House Robber](https://github.com/doocs/leetcode/tree/master/solution/198.House%20Robber) | `Dynamic Programming` |
 | 203 | [Remove Linked List Elements](https://github.com/doocs/leetcode/tree/master/solution/203.Remove%20Linked%20List%20Elements) | `Linked List` |
 | 231 | [Power of Two](https://github.com/doocs/leetcode/tree/master/solution/231.Power%20of%20Two) | `Math`, `Bit Manipulation` |
+| 234 | [Palindrome Linked List](https://github.com/doocs/leetcode/tree/master/solution/234.Palindrome%20Linked%20List) | `Linked List`, `Two Pointers` |
 | 235 | [Lowest Common Ancestor of a Binary Search Tree](https://github.com/doocs/leetcode/tree/master/solution/235.Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree) | `Tree` |
 | 237 | [Delete Node in a Linked List](https://github.com/doocs/leetcode/tree/master/solution/237.Delete%20Node%20in%20a%20Linked%20List) | `Linked List` |
 | 344 | [Reverse String](https://github.com/doocs/leetcode/tree/master/solution/344.Reverse%20String) | `Two Pointers`, `String` |

solution/234.Palindrome Linked List/README.md

@@ -0,0 +1,72 @@
+## 回文链表
+### 题目描述
+
+
+请判断一个链表是否为回文链表。
+
+**示例 1:**
+```
+输入: 1->2
+输出: false
+```
+
+**示例 2:**
+```
+输入: 1->2->2->1
+输出: true
+```
+
+**进阶：**
+
+你能否用 O(n) 时间复杂度和 O(1) 空间复杂度解决此题？
+
+### 解法
+利用快慢指针，找到链表的中间节点，之后对右部分链表进行逆置，然后左右两部分链表依次遍历，若出现对应元素不相等的情况，返回 `false`；否则返回 `true`。
+
+
+```java
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public boolean isPalindrome(ListNode head) {
+        if (head == null || head.next == null) {
+            return true;
+        }
+        ListNode slow = head;
+        ListNode fast = head;
+        while (fast != null && fast.next != null) {
+            slow = slow.next;
+            fast = fast.next.next;
+        }
+        if (fast != null) {
+            slow = slow.next;
+        }
+        
+        ListNode rightPre = new ListNode(-1);
+        while (slow != null) {
+            ListNode t = slow.next;
+            slow.next = rightPre.next;
+            rightPre.next = slow;
+            slow = t;
+        }
+        
+        ListNode p1 = rightPre.next;
+        ListNode p2 = head;
+        while (p1 != null) {
+            if (p1.val != p2.val) {
+                return false;
+            }
+            p1 = p1.next;
+            p2 = p2.next;
+        }
+        return true;
+        
+    }
+}
+```

solution/234.Palindrome Linked List/Solution.java

@@ -0,0 +1,44 @@
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public boolean isPalindrome(ListNode head) {
+        if (head == null || head.next == null) {
+            return true;
+        }
+        ListNode slow = head;
+        ListNode fast = head;
+        while (fast != null && fast.next != null) {
+            slow = slow.next;
+            fast = fast.next.next;
+        }
+        if (fast != null) {
+            slow = slow.next;
+        }
+        
+        ListNode rightPre = new ListNode(-1);
+        while (slow != null) {
+            ListNode t = slow.next;
+            slow.next = rightPre.next;
+            rightPre.next = slow;
+            slow = t;
+        }
+        
+        ListNode p1 = rightPre.next;
+        ListNode p2 = head;
+        while (p1 != null) {
+            if (p1.val != p2.val) {
+                return false;
+            }
+            p1 = p1.next;
+            p2 = p2.next;
+        }
+        return true;
+        
+    }
+}