题目

找出单链表中倒数第k个元素，例如给定1>2>3>4>5>6>7，则单链表中的倒数第k=3个元素为5.

#%%
class LHead:
    def __init__(self):
        self.next = None
    
    def get_ordered_list(self): # 以列表形式打印顺序表
        order_list = []
        if self.next == None or self == None:
            return order_list
        else:
            cur = self.next
            while cur != None:
                order_list.append(cur.data)
                cur = cur.next
            return order_list
        
    def ordered_list_size(self): #获得链表长度
        
        return len(self.get_ordered_list())
    
    def create_ordered_list(self,size): #在新建头结点后面生成有序链表
        i = 0
        tmp = None
        cur = self
        ordered_list = []
         #构造单链表
        while i < size:
             tmp = LNode(i)
             tmp.next = None
             cur.next = tmp
             cur = tmp
             i += 1
        cur = self.next
        while cur != None:
            ordered_list.append(cur.data)
            cur = cur.next
        return ordered_list,self
    
#%%
class LNode:
    def __init__(self,data):
        self.data = data
        self.next = None
        

#%%
def find_kth_element(head,k):
    if head == None or head.next == None:
        return None
    else:
        slow = head.next
        fast = head.next
        for i in range(k): # 可以用while同时判断i < k和 fast!= None
            if fast != None:
                fast = fast.next
            else:
                if i == k :
                    return slow.data
                else:
                    return None
        while fast!= None:
            fast = fast.next
            slow = slow.next
        return slow.data