Leetcode 19. Remove Nth Node From End of List

Description: Given the head of a linked list, remove the nth node from the end of the list and return its head. Follow up: Could you do this in one pass?

Link: https://leetcode.com/problems/remove-nth-node-from-end-of-list/

解题思路: 删除倒数第n个元素，且只遍历一遍。我们熟悉的情况是删除第m个元素，但如果我们知道链表的总长度length，问题就是删除正向第m = length-n+1个元素。为了只遍历一次，在计算length的同时，记录每个node，就可以在一次遍历结束后找到第m-1个元素，然后删除它后面的一个元素，将剩余重新连接。同时考虑删除head的特殊情况。

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        if not head: return head
        p = head
        length = 0
        nodes = []
        while p:
            nodes.append(p)
            p = p.next
            length += 1
        if n == length:
            return head.next
        else:
            pre_node = nodes[length - n - 1]
            pre_node.next = pre_node.next.next
            return head

日期： 2020-11-14  Hope the prepare will go well, hope I could make progress and have large improvement.