技術標籤：LeetCode演算法與資料結構leetcodelinked listtwo pointers

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：我寫的，雙指標。由於是單向連結串列而不是雙向連結串列，其中一指標要反覆橫跳。時雜O(N²)，空雜O(1)。

方法二：使用棧。時雜O(N)，空雜O(N)。

方法三：遞迴。時雜O(N)，空雜O(N)。

方法四：雙指標，反轉連結串列的半部分，這會修改原連結串列結構。時雜O(N)，空雜O(1)。

Submission

import java.util.LinkedList;

import com.lun.util.SinglyLinkedList.ListNode;
 


public class PalindromeLinkedList {

	// 方法一：我寫的
	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：使用棧
	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：遞迴
	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：雙指標，以及要反轉連結串列
	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)));
	}
}