链表问题：链表中倒数第ｋ个节点

思路：
1.是先访问一便链表，知道链表长度ｎ，返回第ｎ－ｋ个结点
2.注意：链表长度小于k

查找倒数第一个节点：

查找倒数第2个节点

代码

class Solution:
    def FindKthToTail(self, head, k):
        if not head:
            return None
        n=0
        t = head
        while(head):
            head = head.next
            n=n+1
        if k>n:
            return None
        i=0
        mm=n-k
        for j in range(n-k):#5-1=4 >>[0~3]4次
            t = t.next
        return t
        
nodeNew=Node_l()
nodeNew.tail_insert(5)
nodeNew.tail_insert(6)
nodeNew.tail_insert(7)
nodeNew.tail_insert(8)
nodeNew.tail_insert(9)

nodeNew.travel()
s=Solution()
s.FindKthToTail(nodeNew.head,2)#5,6,7

链表

class Node(object):
    def __init__(self,data,next=None):
        self.data=data
        self.next=next
        
class Node_l(object):
    def __init__(self):
        self.head=None
    def head_insert(self,data):
        node=Node(data)
        node.next=self.head#赋值 引用
        self.head=node

    def tail_insert(self,data):
        node=Node(data)
        if self.head==None:
            self.head=node
        else:
            cur=self.head
            while cur.next!=None:
                cur=cur.next
            cur.next=node
            # print
            
    def travel(self):
        if self.head ==None:
            return ' '
        cur=self.head
        while cur!=None:
            print (cur.data)
            cur=cur.next
        return ' '