[Leetcode] 19. Remove Nth Node From End of List

Problem:

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.

Idea:

1. 用空间换时间方法。从head开始遍历单向链表，把每一项放入数组中（此时存放的都是链表节点的引用）。遍历完毕后，计算数组的大小（即单向链表的长度），然后用总长度减去n，可得需要移除的项的下标。然后直接访问数组中该下标，修改对应以及临近节点的值即可。

2. n ahead。设置两个游标，其中fast游标比slow游标快n个节点。当fast游标到达链表末端时，slow游标所指的节点就是待删节点。

3. 递归法。先通过递归调用index()函数找到末节点的值，然后把倒数第n个节点之前的节点值往后挪一个位置：“

node.next.val = node.val

”以下是摘抄自作者：

Value-Shifting - AC in 64 ms

My first solution is “cheating” a little. Instead of really removing the nth node, I remove the nth value. I recursively determine the indexes (counting from back), then shift the values for all indexes larger than n, and then always drop the head.

Solution:

1.

class Solution(object):
def removeNthFromEnd(self, head, n):
nodelist = []
if head == None:
return
tmpnode = head
while tmpnode.next  != None:
nodelist.append(tmpnode)
tmpnode = tmpnode.next
nodelist.append(tmpnode)
lenlist = len(nodelist)
targetindex = lenlist - n
if targetindex == 0:
if lenlist == 1:
return None
else:
head = nodelist[1]
elif targetindex == lenlist - 1:
nodelist[lenlist-2].next = None
else:
nodelist[targetindex-1].next = nodelist[targetindex+1]
return head

2.

class Solution:
def removeNthFromEnd(self, head, n):
fast = slow = head
for _ in range(n):
fast = fast.next
if not fast:
return head.next
while fast.next:
fast = fast.next
slow = slow.next
slow.next = slow.next.next
return head

3.

class Solution:
def removeNthFromEnd(self, head, n):
def index(node):
if not node:
return 0
i = index(node.next) + 1
if i > n:
node.next.val = node.val
return i
index(head)
return head.next