Topic
- Linked List
- Two Pointers
Description
https://leetcode.com/problems/palindrome-linked-list/
Given a singly linked list, determine if it is a palindrome.
Example 1:
Input: 1->2
Output: false
Example 2:
Input: 1->2->2->1
Output: true
Follow up:
Could you do it in O(n) time and O(1) space?
Analysis
方法一：我写的，双指针。由于是单向链表而不是双向链表，其中一指针要反复横跳。时杂O(N²)，空杂O(1)。
方法二：使用栈。时杂O(N)，空杂O(N)。
方法三：递归。时杂O(N)，空杂O(N)。
方法四：双指针，反转链表的半部分，这会修改原链表结构。时杂O(N)，空杂O(1)。
Submission
import java.util.LinkedList;
import com.lun.util.SinglyLinkedList.ListNode;
public class PalindromeLinkedList {
// 方法一：我写的
public boolean isPalindrome1(ListNode head) {
if (head == null)
return false;
ListNode p1 = head, p2 = head;
int count = 0;
while (p1 != null) {
p1 = p1.next;
count++;
}
int half = count - count / 2;
while (half-- > 0) {
p2 = p2.next;
}
p1 = head;
for (int i = count / 2; i > 0; i--) {
int step = i - 1;
while (step-- > 0) {
p1 = p1.next;
}
if (p1.val != p2.val) {
return false;
}
p1 = head;
p2 = p2.next;
}
return true;
}
// 方法二：使用栈
public boolean isPalindrome2(ListNode head) {
ListNode temp = head;
LinkedList<Integer> stack = new LinkedList<>();
while (temp != null) {
stack.push(temp.val);
temp = temp.next;
}
while (head != null) {
if (head.val != (int) stack.pop()) {
return false;
}
head = head.next;
}
return true;
}
// 方法三：递归
ListNode ref;
public boolean isPalindrome3(ListNode head) {
ref = head;
return check(head);
}
public boolean check(ListNode node) {
if (node == null)
return true;
boolean ans = check(node.next);
boolean isEqual = (ref.val == node.val) ? true : false;
ref = ref.next;
return ans && isEqual;
}
// 方法四：双指针，以及要反转链表
public boolean isPalindrome4(ListNode head) {
ListNode fast = head, slow = head;
while (fast != null && fast.next != null) {
fast = fast.next.next;
slow = slow.next;
}
if (fast != null) {
// odd nodes: let right half smaller
slow = slow.next;
}
slow = reverse(slow);
fast = head;
while (slow != null) {
if (fast.val != slow.val) {
return false;
}
fast = fast.next;
slow = slow.next;
}
return true;
}
private ListNode reverse(ListNode head) {
ListNode prev = null;
while (head != null) {
ListNode next = head.next;
head.next = prev;
prev = head;
head = next;
}
return prev;
}
}
Test
import static org.junit.Assert.*;
import org.junit.Test;
import com.lun.util.SinglyLinkedList;
public class PalindromeLinkedListTest {
@Test
public void test() {
PalindromeLinkedList obj = new PalindromeLinkedList();
assertFalse(obj.isPalindrome1(SinglyLinkedList.ints2List(1, 2)));
assertTrue(obj.isPalindrome1(SinglyLinkedList.ints2List(1, 2, 2, 1)));
assertFalse(obj.isPalindrome2(SinglyLinkedList.ints2List(1, 2)));
assertTrue(obj.isPalindrome2(SinglyLinkedList.ints2List(1, 2, 2, 1)));
assertFalse(obj.isPalindrome3(SinglyLinkedList.ints2List(1, 2)));
assertTrue(obj.isPalindrome3(SinglyLinkedList.ints2List(1, 2, 2, 1)));
assertFalse(obj.isPalindrome4(SinglyLinkedList.ints2List(1, 2)));
assertTrue(obj.isPalindrome4(SinglyLinkedList.ints2List(1, 2, 2, 1)));
}
}