Title description
Given a linked list, please determine whether the linked list is a palindrome structure.
Example 1
enter
[1,2,2,1]
return value
true
For a palindrome linked list, it can be found that when the pointer node is traversing it, walking to the middle position is equivalent to walking backward from the middle,
such as 1->2->3->2->1, 1-> 2->2->1
So the part after the middle can be regarded as the reversal of the first half,
so the part after the middle can be reversed, and a linked list that is the same as the first half can be obtained, which also becomes the judgment back Text structure conditions
import java.util.*;
/*
* public class ListNode {
* int val;
* ListNode next = null;
* }
*/
public class Solution {
/**
* 对于一个回文链表可以发现,当指针节点在对其进行遍历时,走到中间位置就等同于从中间倒着往前走
* 如 1->2->3->2->1,1->2->2->1
* 所以可以将中间以后的部分视为前半部分的反转
* 因此可以将中间以后的部分进行反转,就能得到一个和前半部分相同的链表,这也变成了判断回文结构的条件
*/
public boolean isPail (ListNode head) {
if(head == null || head.next == null){
return true;
}
ListNode fastNode = head;
ListNode slowNode = head;
while(fastNode != null && fastNode.next != null){
fastNode = fastNode.next.next;
slowNode = slowNode.next;
}
//针对后半段链表 进行反转
ListNode curSlowNode = slowNode.next;
ListNode preSlowNode = slowNode;
ListNode tempNode = null;
while(curSlowNode!=null){
tempNode = curSlowNode.next;
curSlowNode.next = preSlowNode;
preSlowNode = curSlowNode;
curSlowNode = tempNode;
}
slowNode.next = null;
while(preSlowNode!=null){
if(head.val != preSlowNode.val){
return false;
}
head = head.next;
preSlowNode = preSlowNode.next;
}
return true;
}
}