【Hot100】LeetCode—234. 回文链表

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• 原题连接：234. 回文链表

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1- 思路

快慢指针+链表拆分+反转链表

思路
①将链表拆分前后两个部分——>找拆分点、②反转后面部分、③根据反转结果，同时利用两个指针遍历

• ① 找拆分点：快慢指针
  • 利用快慢指针，满指针的 next 就是 后半部分的头指针
• ② 反转链表
  • 递归反转后半部分
• ③ 遍历判断
  • 依次同时移动两个指针判断

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2- 实现

⭐234. 回文链表——题解思路

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
class Solution {public boolean isPalindrome(ListNode head) {ListNode endA = endOfA(head);boolean res = true;ListNode headB = reverseL(endA.next);ListNode curA = head;ListNode curB = headB;while(res && curB!=null){if(curA.val != curB.val){res = false;}curA = curA.next;curB = curB.next;}// 恢复 B reverseL(headB);return res;}public ListNode endOfA(ListNode head){ListNode slow = head;ListNode fast = head;while(fast.next!=null && fast.next.next!=null){slow = slow.next;fast = fast.next.next;}return slow;}public ListNode reverseL(ListNode head){if(head==null || head.next==null){return head;}ListNode cur = reverseL(head.next);head.next.next = head;head.next = null;return cur;}
}

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3- ACM 实现

public class isPalindrome {public static class ListNode {int val;ListNode next;ListNode(int x) {val = x;next = null;}}public static boolean isP(ListNode head){ListNode endA = endOfA(head);// 反转ListNode headB = reverseL(endA.next);ListNode curA = head;ListNode curB = headB;boolean res = true;while(curB!=null){if(curA.val != curB.val){res = false;}curA = curA.next;curB = curB.next;}return res;}private static ListNode endOfA(ListNode head){ListNode slow = head;ListNode fast = head;while(fast.next!=null && fast.next.next!=null){slow = slow.next;fast = fast.next;}return slow;}private static ListNode reverseL(ListNode head){if(head == null || head.next==null){return head;}ListNode cur = reverseL(head.next);head.next.next = head;head.next = null;return cur;}public static void main(String[] args) {Scanner sc = new Scanner(System.in);System.out.println("输入链表长度");int n = sc.nextInt();ListNode head = null,tail=null;for(int i = 0 ; i < n;i++){ListNode nowNode  = new ListNode(sc.nextInt());if(head==null){head = nowNode;tail = nowNode;}else{tail.next = nowNode;tail = nowNode;}}System.out.println("结果是"+isP(head));}
}