Description

Given a linked list, remove the nth node from the end of list and return its head.

For example,

Given linked list: 1->2->3->4->5, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5.
Note:
Given n will always be valid.
Try to do this in one pass.

給定一個連結串列，刪除連結串列的倒數第 n 個節點並返回頭結點。

例如，

給定一個連結串列: 1->2->3->4->5, 並且 n = 2.

當刪除了倒數第二個節點後連結串列變成了 1->2->3->5.

說明:

給的 n 始終是有效的。

嘗試一次遍歷實現。

Solution

構建雙指標p1與p2，p1先走n步，然後一同運動，當p1指向表尾，p2指向的next即是倒數第N個節點，刪除即可。

# -*- coding: utf-8 -*-
"""
Created on Sat Mar 17 16:35:29 2018

@author: Saul
"""
 


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

class Solution:
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        dummy = ListNode(0
 
)
        dummy.next = head
        p1 = p2 = dummy
        for i in range(n):
            p1 = p1.next
        while p1.next:
            p1 = p1.next
            p2 = p2.next
        p2.next = p2.next.next
        return dummy.next