Palindrome Linked List

Author: w2xi

README.md

@@ -15,6 +15,7 @@
 |141|[环形链表](https://leetcode-cn.com/problems/linked-list-cycle/)|[JavaScript](./algorithms/linked-list-cycle.js)|Easy|
 |189|[轮转数组](https://leetcode-cn.com/problems/rotate-array/)|[JavaScript](./algorithms/rotate-array.js)|Medium|
 |206|[反转链表](https://leetcode-cn.com/problems/reverse-linked-list/)|[JavaScript](./algorithms/reverse-linked-list.js)|Easy|
+|234|[回文链表](https://leetcode-cn.com/problems/palindrome-linked-list/)|[JavaScript](./algorithms/palindrome-linked-list.js)|Easy|
 |344|[反转字符串](https://leetcode-cn.com/problems/reverse-string/)|[JavaScript](./algorithms/reverse-string.js)|Easy|
 |349|[两个数组的交集](https://leetcode-cn.com/problems/intersection-of-two-arrays/)|[JavaScript](./algorithms/intersection-of-two-arrays.js)|Easy|
 |350|[两个数组的交集 II](https://leetcode-cn.com/problems/intersection-of-two-arrays-ii/)|[JavaScript](./algorithms/intersection-of-two-arrays-ii.js)|Easy|

algorithms/palindrome-linked-list.js

@@ -0,0 +1,74 @@
+// 方法一：栈的后进先出
+
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val, next) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.next = (next===undefined ? null : next)
+ * }
+ */
+/**
+ * 回文链表
+ * @param {ListNode} head
+ * @return {boolean}
+ */
+var isPalindrome = function (head) {
+  let curr = head;
+  let arr = [];
+
+  while (curr) {
+    arr.push(curr.val);
+    curr = curr.next;
+  }
+
+  while (head) {
+    if (arr.pop() !== head.val) {
+      return false;
+    }
+    head = head.next;
+  }
+
+  return true;
+};
+
+
+// 快慢节点 + 反转链表
+
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val, next) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.next = (next===undefined ? null : next)
+ * }
+ */
+/**
+ * @param {ListNode} head
+ * @return {boolean}
+ */
+var isPalindrome = function (head) {
+  let slow = head,
+    fast = head,
+    prev = null;
+  // 快慢指针找到 middle node
+  while (fast) {
+    slow = slow.next;
+    fast = fast.next ? fast.next.next : fast.next;
+  }
+  // 反转后半部分链表
+  while (slow) {
+    let temp = slow.next;
+    slow.next = prev;
+    prev = slow;
+    slow = temp;
+  }
+  // 依次比较
+  while (head && prev) {
+    if (head.val !== prev.val) {
+      return false;
+    }
+    head = head.next;
+    prev = prev.next;
+  }
+
+  return true;
+};