题解 | #链表中倒数最后k个结点#

# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param pHead ListNode类 
# @param k int整型 
# @return ListNode类
#
class Solution:
    def FindKthToTail(self , pHead: ListNode, k: int) -> ListNode:
        # write code here
        cur = pHead
        if not cur:
            return None
        length = 0
#解法1：将链表所有数值全部取到列表中，然后把-k到-1的列表元素提取出来新建一个链表返回。
#         list1 = []
#         pre = post = ListNode(0)
#         while cur:
#             length += 1
#             list1.append(cur.val)
#             cur = cur.next
#         if length < k:
#             return None
#         i = -k
#         while -k <= i and i <= -1:
#             post.next = ListNode(0)
#             post.next.val = list1[i]
#             post = post.next
#             i += 1
#         return pre.next
#解法2：先遍历链表长度，当遍历长度length-k次后，然后当前指针指向的子链表即可
        while cur:
            length = length + 1
            cur = cur.next
        if length < k:
            return None
        a = length - k
        if a == 0:
            return pHead
        cur = pHead
        i = 0 
        while cur:
            i = i + 1
            cur = cur.next
            if a == i:
                return cur