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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
方法一:三指针
Python3
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:def reverseList(self, head: ListNode) -> ListNode:pre,cur=None,headwhile cur:nxt=cur.nextcur.next=prepre=curcur=nxtreturn pre
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) {ListNode pre = null, p = head;while (p != null) {ListNode q = p.next;p.next = pre;pre = p;p = q;}return pre;}
}
C++
/*** 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;}
};
Go
/*** Definition for singly-linked list.* type ListNode struct {* Val int* Next *ListNode* }*/
func reverseList(head *ListNode) *ListNode {var pre *ListNodefor p := head; p != nil; {q := p.Nextp.Next = prepre = pp = q}return pre
}
JavaScript
/*** 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;
};
C#
/*** 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;}
}
Swift
/*** Definition for singly-linked list.* public class ListNode {* public var val: Int* public var next: ListNode?* public init(_ val: Int) {* self.val = val* self.next = nil* }* }*/class Solution {func reverseList(_ head: ListNode?) -> ListNode? {var prev: ListNode? = nilvar current = headwhile current != nil {let next = current?.nextcurrent?.next = prevprev = currentcurrent = next}return prev}
}
方法二:递归
1）res = reverseList(head.next): 作用体现在框里（抽象）
2）head.next.next=head,head.next=none:体现在最前一段反转
python
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:def reverseList(self, head: ListNode) -> ListNode:#返回反转后的链表的头节点if not head or not head.next:return headres=self.reverseList(head.next)head.next.next=headhead.next=Nonereturn res
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 res = reverseList(head.next);head.next.next = head;head.next = null;return res;}
}
