1. 206. 反轉連結串列
輸入：head = [1,2,3,4,5]
輸出：[5,4,3,2,1]

class Solution(object):
    def reverseList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """

        pre = None
        cur = head
        while cur:
            tmp = cur.next
            cur.next = pre
            pre 
 
= cur
            cur = tmp 
        return pre

2. 劍指 Offer 22. 連結串列中倒數第k個節點 ttps://leetcode-cn.com/problems/reverse-linked-list

class Solution(object):
    def getKthFromEnd(self, head, k):
        """
        :type head: ListNode
        :type k: int
        :rtype: ListNode
        """
　　　　　#雙指標，faster比slower快k個node
 

        former, latter = head, head

        for i in range(k):
            former = former.next
        while former:
            former, latter = former.next, latter.next
        
        return latter

3. 876. 連結串列的中間結點

class Solution(object):
    def middleNode(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        
 
"""
　　　　　# 雙指標，faster是slower的兩倍

        faster = slower = head

        while faster and faster.next:
            slower = slower.next
            faster = faster.next.next

        return slower
        

        l1, l2 = head, head
        count = 0

        while l1:
            count += 1
            l1 = l1.next # count the length of linkedlist
        mid = count/2  # get the mid num 

        for i in range(mid):
            l2 = l2.next

        return l2