Leecode刷题之旅-C语言/python-206反转链表

/*

 * @lc app=leetcode.cn id=206 lang=c

 *

 * [206] 反转链表

 *

 * https://leetcode-cn.com/problems/reverse-linked-list/description/

 *

 * algorithms

 * Easy (58.89%)

 * Total Accepted:    41.2K

 * Total Submissions: 69.9K

 * Testcase Example:  '[1,2,3,4,5]'

 *

 * 反转一个单链表。

 *

 * 示例:

 * 输入: 1->2->3->4->5->NULL

 * 输出: 5->4->3->2->1->NULL

 *

 * 进阶:

 * 你可以迭代或递归地反转链表。你能否用两种方法解决这道题？

 *

 */

/**

 * Definition for singly-linked list.

 * struct ListNode {

 *     int val;

 *     struct ListNode *next;

 * };

 */

struct ListNode* reverseList(struct ListNode* head) {

    struct ListNode* p = head;

    struct ListNode *q = NULL;

    if(head==NULL||head->next==NULL)

    {

        return head;

    }

    while(p!=NULL){

        struct ListNode *temp;

        temp = p->next;

        p->next=q;

        q = p;

        p = temp;

    }

    return q;

}

这里设置了三个listnode指针变量。 如果比较难理解的话画个图就会好懂很多。

-----------------------------------------------------------------------------------------------------

python:

#

# @lc app=leetcode.cn id=206 lang=python3

#

# [206] 反转链表

#

# https://leetcode-cn.com/problems/reverse-linked-list/description/

#

# algorithms

# Easy (58.89%)

# Total Accepted:    41.2K

# Total Submissions: 69.9K

# Testcase Example:  '[1,2,3,4,5]'

#

# 反转一个单链表。

#

# 示例:

#

# 输入: 1->2->3->4->5->NULL

# 输出: 5->4->3->2->1->NULL

#

# 进阶:

# 你可以迭代或递归地反转链表。你能否用两种方法解决这道题？

#

#

# 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:

        if head is None:

            return None

        cur = head

        pre = None

        nxt = cur.next

        while nxt:

            cur.next = pre

            pre = cur

            cur = nxt

            nxt = nxt.next

        cur.next = pre

        head = cur

        return head