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。

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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