1、从头到尾打印链表

输入一个链表，按链表值从尾到头的顺序返回一个ArrayList。

# -*- coding:utf-8 -*-class ListNode:     def __init__(self, x):         self.val = x         self.next = Noneclass Solution:    def printListFromTailToHead(self, listNode):        l =[]        while listNode:            l.append(listNode.val)            listNode = listNode.next        return l[::-1]

2、链表中倒数第k个节点

输入一个链表，输出该链表中倒数第k个结点。

# -*- coding:utf-8 -*-class ListNode:    def __init__(self, x):        self.val = x        self.next = Noneclass Solution:    def FindKthToTail(self, head, k):        node_list = []        while head:            node_list.append(head)            head = head.next        if k < 1 or k > len(node_list):            return        return node_list[-k]

3、反转链表

输入一个链表，反转链表后，输出新链表的表头。

# -*- coding:utf-8 -*-class ListNode:    def __init__(self, x):        self.val = x        self.next = None        class Solution:    def ReverseList(self, pHead):        if pHead is None or pHead.next is None:            return pHead        pre = None        cur = pHead        while cur:            temp = cur.next            cur.next = pre            pre = cur            cur = temp        return pre

 

待续...