python實現 LeetCode19——Remove Nth Node From End of List

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.

這個題的思路就是用兩個指針p和q指向head，然後q向後移動n步，進入循環，p和q同時向後移動，直到q.next爲空，去掉p後面的節點。還有就是對於q爲空，q.next不存在進行特殊討論下。

leetcode提交代碼

class Solution():
    def removeNthFromEnd(self, head, n):
        p=head
        q=head
        while n!=0:
            q=q.next
            n=n-1
            if q==None :
                return head.next
        while q.next!=None:
            p=p.next
            q=q.next
        r=p.next
        p.next=r.next
        return head

另外附上可以檢驗的代碼

# -*- coding: utf-8 -*
class Node(object):
    def __init__(self,val,p=None):
        self.data = val
        self.next = p

class list(object):
    def __init__(self):                 #頭節點爲0
        self.head = 0
    def initlist(self,data):
        self.head = Node(data[0])
        p = self.head
        for i in data[1:]:
            node = Node(i)
            p.next = node
            p = p.next

class Solution():
    def removeNthFromEnd(self, head, n):
        p=head
        q=head
        while n!=0:
            q=q.next
            n=n-1
            if q==None :
                return head.next
        while q.next!=None:
            p=p.next
            q=q.next
        r=p.next
        p.next=r.next
        return head
    def shuchu(self,head):
        p=head
        while p!=None:
            print p.data
            p=p.next
s = Solution()
l=list()
l.initlist([1,2,3,4])
p=s.removeNthFromEnd(l.head,3)
s.shuchu(p)