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/

解題思路: 刪除倒數第n個元素，且只遍歷一遍。我們熟悉的情況是刪除第m個元素，但如果我們知道連結串列的總長度length，問題就是刪除正向第m = length-n+1個元素。為了只遍歷一次，在計算length的同時，記錄每個node，就可以在一次遍歷結束後找到第m-1個元素，然後刪除它後面的一個元素，將剩餘重新連線。同時考慮刪除head的特殊情況。

# 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.