本问题对应的 leetcode 原文链接：剑指 Offer 24. 反转链表
问题描述
定义一个函数，输入一个链表的头节点，反转该链表并输出反转后链表的头节点。
示例 1：
输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL
限制：
0 <= 节点个数 <= 5000
解题思路
视频讲解直达： 本题视频讲解
代码实现
class Solution {
// 原地反转
public ListNode reverseList(ListNode head) {
if(head == null || head.next == null){
return head;
}
ListNode cur = head, pre = null;
while(cur != null){
ListNode temp = cur.next;
cur.next = pre;
pre = cur;
cur = temp;
}
return pre;
// 额外空间复杂度=1
}
// 递归
public ListNode reverseList(ListNode head) {
if(head == null || head.next == null){
return head;
}
// 反转子链表
ListNode temp = reverseList(head.next);
head.next.next = head;
head.next = null;
return temp;
}
}
Python
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def reverseList(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
if head is None or head.next is None:
return head
cur = head
pre = None
while cur is not None:
temp = cur.next
cur.next = pre
pre = cur
cur = temp
return pre
C++
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* reverseList(ListNode* head) {
if (head == nullptr || head->next == nullptr) {
return head;
}
ListNode* cur = head;
ListNode* pre = nullptr;
while (cur != nullptr) {
ListNode* temp = cur->next;
cur->next = pre;
pre = cur;
cur = temp;
}
return pre;
}
};
Go
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func reverseList(head *ListNode) *ListNode {
if head == nil || head.Next == nil {
return head
}
var cur *ListNode = head
var pre *ListNode = nil
for cur != nil {
temp := cur.Next
cur.Next = pre
pre = cur
cur = temp
}
return pre
}
JS
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} head
* @return {ListNode}
*/
var reverseList = function(head) {
if (head == null || head.next == null) {
return head;
}
let cur = head,
pre = null;
while (cur != null) {
let temp = cur.next;
cur.next = pre;
pre = cur;
cur = temp;
}
return pre;
// 额外空间复杂度=1
// 递归
// if(head == null || head.next == null){
// return head;
// }
// // 反转子链表
// let temp = reverseList(head.next);
// head.next.next = head;
// head.next = null;
// return temp;
};
shuaidi