题目:

编写一个函数,检查输入的链表是否是回文的。

 

示例 1:

输入: 1->2

输出: false

示例 2:

输入: 1->2->2->1

输出: true

代码实现:

class Solution {
public boolean isPalindrome(ListNode head) {
if (head == null) {
return true;
}

// 找到前半部分链表的尾节点并反转后半部分链表
ListNode firstHalfEnd = endOfFirstHalf(head);
ListNode secondHalfStart = reverseList(firstHalfEnd.next);

// 判断是否回文
ListNode p1 = head;
ListNode p2 = secondHalfStart;
boolean result = true;
while (result && p2 != null) {
if (p1.val != p2.val) {
result = false;
}
p1 = p1.next;
p2 = p2.next;
}

// 还原链表并返回结果
firstHalfEnd.next = reverseList(secondHalfStart);
return result;
}

private ListNode reverseList(ListNode head) {
ListNode prev = null;
ListNode curr = head;
while (curr != null) {
ListNode nextTemp = curr.next;
curr.next = prev;
prev = curr;
curr = nextTemp;
}
return prev;
}

private ListNode endOfFirstHalf(ListNode head) {
ListNode fast = head;
ListNode slow = head;
while (fast.next != null && fast.next.next != null) {
fast = fast.next.next;
slow = slow.next;
}
return slow;
}
}