19. Remove Nth Node From End of List

题目:

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.

链接:  http://leetcode.com/problems/remove-nth-node-from-end-of-list/

一刷

Python

删除节点一般需要留指向被删除节点前一个节点。这类题目经常每换一步都要validate，比如开始查n＝＝0，之后range(n-1)的循环里要检查fast == None，循环之后要检查是否可直接退出，第二个循环里检查fast.next，最后检查prev的指向内容。

这道题很简单，可是要按顺序在每一步找到合适的validate还是有些麻烦，最好是从一般到特殊来检查是否需要额外添加。测试例子1-->2-->3可以用n从1-4来测试。

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

class Solution(object):
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        if n == 0:
            return head
        fast = head
        for i in range(n-1):
            if fast != None:
                fast = fast.next
            else:
                return head
        else:
            if fast == None:
                return head
        prev = None
        slow = head
        while fast.next:
            fast = fast.next
            prev = slow
            slow = slow.next
        if prev:
            prev.next = slow.next
            return head
        else:
            return head.next

好思路：建立dummy节点，省略很多边界条件的检查，n>0时只移动fast，n<=0时同时移动fast, slow。

class Solution(object):
    def removeNthFromEnd(self, head, n):
        if n == 0:
            return head
        dummy = ListNode(0)
        dummy.next = head
        fast = dummy
        slow = dummy
        
        while fast.next:
            fast = fast.next
            if n <= 0:
                slow = slow.next
            n -= 1
        slow.next = slow.next.next
        return dummy.next