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