234. 回文链表
难度简单 872
请判断一个链表是否为回文链表。
示例 1:
输入: 1->2
输出: false
示例 2:
输入: 1->2->2->1
输出: true
进阶：
你能否用 O(n) 时间复杂度和 O(1) 空间复杂度解决此题？
/**
* 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) {
List<Integer> vals = new ArrayList<Integer>();
// 将链表的值复制到数组中
ListNode currentNode = head;
while (currentNode != null) {
vals.add(currentNode.val);
currentNode = currentNode.next;
}
// 使用双指针判断是否回文
int front = 0;
int back = vals.size() - 1;
while (front < back) {
if (!vals.get(front).equals(vals.get(back))) {
return false;
}
front++;
back--;
}
return true;
}
// 作者：LeetCode-Solution
// 链接：https://leetcode-cn.com/problems/palindrome-linked-list/solution/hui-wen-lian-biao-by-leetcode-solution/
}栈
/**
* 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) {
Stack<Integer> stk =new Stack();
if (null==head || null==head.next) return true;
ListNode p = head;
while (p != null) {
stk.push(p.val);
p = p.next;
}
while (head != null){
if(head.val == stk.peek()) {
stk.pop();
} else return false;
head = head.next;
}
return true;
}
// 作者：wo-yao-chu-qu-luan-shuo
// 链接：https://leetcode-cn.com/problems/palindrome-linked-list/solution/hui-wen-lian-biao-fu-zhu-zhan-by-wo-yao-ab2uc/
}
