给你一个单链表的头节点 head
,请你判断该链表是否为。回文链表如果是,返回 true
;否则,返回 false
。
示例 1:
输入:head = [1,2,2,1] 输出:true
示例 2:
输入:head = [1,2] 输出:false
解法: 先利用快慢指针,找到链表中点,然后根据中点翻转后续链表,反转后在跟第一个链表比对
class Solution {
public boolean isPalindrome(ListNode head) {
//首先找到链表的中点,然后反转后半部分,然后再比较前半部分
ListNode slow = head;
ListNode fast = head.next;
while(fast !=null && fast .next != null){
slow = slow.next;
fast = fast.next.next;
}
//翻转后半部分
ListNode second = reverseList(slow.next);
slow.next = null;
//比较前半部分
ListNode first = head;
while(second !=null){
if(first.val == second.val){
first = first.next;
second = second.next;
}else {
return false;
}
}
return true;
}
private ListNode reverseList(ListNode slow) {
ListNode prev = null;
ListNode cur = slow;
ListNode next = null;
while(cur != null){
next = cur.next;
cur.next = prev;
prev = cur;
cur = next;
}
return prev;
}
}