Question

剑指 Offer II 024. 反转链表

题目描述

给定单链表的头节点 head ，请反转链表，并返回反转后的链表的头节点。

 

示例 1：

输入：head = [1,2,3,4,5]
输出：[5,4,3,2,1]

示例 2：

输入：head = [1,2]
输出：[2,1]

示例 3：

输入：head = []
输出：[]

 

提示：

• 链表中节点的数目范围是 [0, 5000]
• -5000 <= Node.val <= 5000

 

进阶：链表可以选用迭代或递归方式完成反转。你能否用两种方法解决这道题？

 

注意：本题与主站 206 题相同： https://leetcode.cn/problems/reverse-linked-list/

解法

方法一

# Definition for singly-linked list.
# class ListNode:
#   def __init__(self, x):
#     self.val = x
#     self.next = None


class Solution:
  def reverseList(self, head: ListNode) -> ListNode:
    pre, p = None, head
    while p:
      q = p.next
      p.next = pre
      pre = p
      p = q
    return pre

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *   int val;
 *   ListNode next;
 *   ListNode(int x) { val = x; }
 * }
 */
class Solution {
  public ListNode reverseList(ListNode head) {
    ListNode pre = null, p = head;
    while (p != null) {
      ListNode q = p.next;
      p.next = pre;
      pre = p;
      p = q;
    }
    return pre;
  }
}

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *   int val;
 *   ListNode *next;
 *   ListNode() : val(0), next(nullptr) {}
 *   ListNode(int x) : val(x), next(nullptr) {}
 *   ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
  ListNode* reverseList(ListNode* head) {
    ListNode* pre = nullptr;
    ListNode* p = head;
    while (p) {
      ListNode* q = p->next;
      p->next = pre;
      pre = p;
      p = q;
    }
    return pre;
  }
};

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *   Val int
 *   Next *ListNode
 * }
 */
func reverseList(head *ListNode) *ListNode {
  var pre *ListNode
  for p := head; p != nil; {
    q := p.Next
    p.Next = pre
    pre = p
    p = q
  }
  return pre
}

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *   this.val = (val===undefined ? 0 : val)
 *   this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var reverseList = function (head) {
  let pre = null;
  for (let p = head; p; ) {
    let q = p.next;
    p.next = pre;
    pre = p;
    p = q;
  }
  return pre;
};

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *   public int val;
 *   public ListNode next;
 *   public ListNode(int val=0, ListNode next=null) {
 *     this.val = val;
 *     this.next = next;
 *   }
 * }
 */
public class Solution {
  public ListNode ReverseList(ListNode head) {
    ListNode pre = null;
    for (ListNode p = head; p != null;)
    {
      ListNode t = p.next;
      p.next = pre;
      pre = p;
      p = t;
    }
    return pre;
  }
}

方法二

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *   int val;
 *   ListNode next;
 *   ListNode(int x) { val = x; }
 * }
 */
class Solution {
  public ListNode reverseList(ListNode head) {
    if (head == null || head.next == null) {
      return head;
    }
    ListNode res = reverseList(head.next);
    head.next.next = head;
    head.next = null;
    return res;
  }
}