'''
Given a linked list, remove the n-th node from the end of list and return its head.
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.
Follow up:
Could you do this in one pass?
'''
# 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
"""
dummy = ListNode(-217)
dummy.next = head
fast = dummy
slow = dummy
for i in range(n + 1):
fast = fast.next
while(fast):
fast = fast.next
slow = slow.next
slow.next = slow.next.next
return dummy.next
Several things to consider:
We actually need to find the (n + 1)-th node from the end for deleting the n-th node.
How many steps fast should move ahead? n + 1 for either having dummy node or not.
Dummy node is necessary for removing the 1st node (the length-th node from end).
