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
Python3 Java C++ Go JavaScript C#
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16 # 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
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20 /**
* 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 ;
}
}
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24 /**
* 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 ;
}
};
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17 /**
* 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
}
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21 /**
* 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 ;
};
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24 /**
* 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
GitHub