查找链表倒数第k个节点

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def getKthFromEnd(self, head: ListNode, k: int) -> ListNode:
        # 快慢指针，快指针先走k步，然后快慢指针一起移动，当快指针为none时，此时慢指针为倒数第k个节点

        fast = slow = head
        for _ in range(k):
            fast = fast.next
        
        while fast:
            fast = fast.next
            slow = slow.next
        
        return slow