算法: 翻转链表 206. Reverse Linked List

Posted 2021-11-10 架构师易筋

206. Reverse Linked List

Given the head of a singly linked list, reverse the list, and return the reversed list.

Example 1:

Input: head = [1,2,3,4,5]
Output: [5,4,3,2,1]

Example 2:

Input: head = [1,2]
Output: [2,1]

Example 3:

Input: head = []
Output: []

Constraints:

• The number of nodes in the list is the range [0, 5000].
• -5000 <= Node.val <= 5000

递归解法

/**
 * 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 ListNode reverseList(ListNode head) {
        return helper(head, null);
    }
	private ListNode helper(ListNode newHead, ListNode head) {
		if (newHead == null) return head;
		ListNode next = newHead.next;
		newHead.next = head;
		return helper(next, newHead);
	}
}

遍历解法

/**
 * 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 ListNode reverseList(ListNode head) {
        ListNode newHead = null;
		while (head != null) {
			ListNode next = head.next;
			head.next = newHead;
			newHead = head;
			head = next;
		}
		return newHead;
    }
}