Reverse Linked List（反转单向链表）

Posted 2020-10-06 人总闲

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

Reverse a singly linked list.

递归方法：递归调用直到最后一个节点再开始反转，注意保存反转后的头结点返回

Java

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) { val = x; }
 7  * }
 8  */
 9 class Solution {
10     public ListNode reverseList(ListNode head) {
11         if(head == null || head.next == null) {
12             return head;
13         }
14         ListNode newHead = reverseList(head.next);
15         head.next.next = head;
16         head.next = null;
17         return newHead;
18     }
19 }

Python

 1 # -*- coding:utf-8 -*-
 2 # 递归
 3 # class ListNode:
 4 #     def __init__(self, x):
 5 #         self.val = x
 6 #         self.next = None
 7 class Solution:
 8     # 返回ListNode
 9     def ReverseList(self, pHead):
10         # write code here
11         if pHead == None or pHead.next == None:
12             return pHead
13         tmp = self.ReverseList(pHead.next)
14         pHead.next.next = pHead
15         pHead.next = None
16         return tmp

 

迭代方法：两个指针从头开始依次反转，注意保存下一次反转的节点指针

Java

/**
 * 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 pre = head, cur = head.next, next = null;
        while(cur != null) {
            next = cur.next;
            cur.next = pre;
            if(pre == head) {
                pre.next = null;
            }
            pre = cur;
            cur = next;
        }
        return pre;
    }
}

Python

# -*- coding:utf-8 -*-
# 迭代
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    # 返回ListNode
    def ReverseList(self, pHead):
        # write code here
        if pHead == None or pHead.next == None:
            return pHead
        pre_node, cur_node = pHead, pHead.next
        while pre_node and cur_node:
            tmp_node = cur_node.next
            cur_node.next = pre_node
            pre_node = cur_node
            cur_node = tmp_node
        pHead.next = None
        return pre_node