文章目录
- 234. 回文链表
- 解题
- 方法一:复制到数组,然后左右指针,对比
- 方法二:反转链表
234. 回文链表
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
提示:
链表中节点数目在范围[1, 105] 内
0 <= Node.val <= 9
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?
解题
方法一:复制到数组,然后左右指针,对比
/**
* 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 {
// 复制到数组,然后左右指针,对比
public boolean isPalindrome(ListNode head) {
List<Integer> dataList = new ArrayList<>();
ListNode curNode = head;
while (curNode != null) {
dataList.add(curNode.val);
curNode = curNode.next;
}
// 左右指针,对比
int left = 0;
int right = dataList.size() - 1;
while (left < right) {
if (!dataList.get(left).equals(dataList.get(right))) {
return false;
}
left++;
right--;
}
return true;
}
}
方法二:反转链表
/**
* 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 {
// 快慢指针,反转链表, 之后对比,最后还原链表
public boolean isPalindrome(ListNode head) {
boolean result = true;
if (head == null) {
return result;
}
// 找到前半段的最后节点
ListNode firstHalfLastNode = endOfFirstHalf(head);
// 反转后半段链表,并返回头节点
ListNode secondHalfStart = reverseList(firstHalfLastNode.next);
// 对比
ListNode p1 = head;
ListNode p2 = secondHalfStart;
while(result && p2 != null) {
if (p1.val != p2.val) {
result = false;
}
p1 = p1.next;
p2 = p2.next;
}
// 还原链表,返回结果
firstHalfLastNode.next = reverseList(secondHalfStart);
return result;
}
// 反转链表(借用两个指针)
private ListNode reverseList(ListNode head) {
ListNode pre = null;
ListNode cur = head;
while (cur != null) {
// 复制一份
ListNode temp = cur.next;
// 反转指针
cur.next = pre;
// 指针前移
pre = cur;
cur = temp;
}
return pre;
}
// 找到前半段的最后节点(快慢指针)
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;
}
}