LeetCode - Easy - 206. Reverse Linked List

Topic

• Linked List

Description

https://leetcode.com/problems/reverse-linked-list/

Reverse a singly linked list.

Example:

Input: 1->2->3->4->5->NULL
Output: 5->4->3->2->1->NULL

Follow up:

A linked list can be reversed either iteratively or recursively. Could you implement both?

Analysis

方法一：迭代法。

方法二：別人寫的迭代法。

方法三：遞迴法。

Submission

import com.lun.util.SinglyLinkedList.ListNode;

public class ReverseLinkedList {

	// 方法一：
	public ListNode reverseList1(ListNode head) {
		ListNode p1 = null, p2 = head, p3 = head.next;

		while (p2 != null) {
			p2.next = p1;
			p1 = p2;
			p2 = p3;
			if (p3 != null)
				p3 = p3.next;
		}
		return p1;
	}

	// 方法二：
	public ListNode reverseList2(ListNode head) {
		/* iterative solution */
		ListNode newHead = null;
		while (head != null) {
			ListNode next = head.next;
			head.next = newHead;
			newHead = head;
			head = next;
		}
		return newHead;
	}

	// 方法三：
	public ListNode reverseList3(ListNode head) {
		/* recursive solution */
		return reverseListInt(head, null);
	}

	private ListNode reverseListInt(ListNode head, ListNode newHead) {
		if (head == null)
			return newHead;
		ListNode next = head.next;
		head.next = newHead;
		return reverseListInt(next, head);
	}
}

Test

import com.lun.util.SinglyLinkedList.ListNode;

public class ReverseLinkedList {

	// 方法一：
	public ListNode reverseList1(ListNode head) {
		ListNode p1 = null, p2 = head, p3 = head.next;

		while (p2 != null) {
			p2.next = p1;
			p1 = p2;
			p2 = p3;
			if (p3 != null)
				p3 = p3.next;
		}
		return p1;
	}

	// 方法二：
	public ListNode reverseList2(ListNode head) {
		/* iterative solution */
		ListNode newHead = null;
		while (head != null) {
			ListNode next = head.next;
			head.next = newHead;
			newHead = head;
			head = next;
		}
		return newHead;
	}

	// 方法三：
	public ListNode reverseList3(ListNode head) {
		/* recursive solution */
		return reverseListInt(head, null);
	}

	private ListNode reverseListInt(ListNode head, ListNode newHead) {
		if (head == null)
			return newHead;
		ListNode next = head.next;
		head.next = newHead;
		return reverseListInt(next, head);
	}

}