LCR 027. 回文链表
给定一个链表的 头节点 head ，请判断其是否为回文链表。
如果一个链表是回文，那么链表节点序列从前往后看和从后往前看是相同的。
示例 1：
输入: head = [1,2,3,3,2,1] 输出: true
思路：
反转后半截 + 快慢指针找到后半截起点【整个反转再同时遍历——>超时】
【快慢指针找中点——>快指针一次走两步、慢指针一次走一步】
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;}
【完整代码 来源力扣官方】
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;} }
)
