Remove Nth Node From End of List

Similar Problems:

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.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

## Blog link: https://code.dennyzhang.com/remove-nth-node-from-end-of-list ## Basic Ideas: Two pointers with distance of n+1 ## Since the head node might be deleted, we need a dummy node ## One pass ## 1->2->3->4->5, n=2 ## p . . q ## ## Assumption: If n is bigger than the length of the list, do nothing ## Complexity: Time O(n), Space O(1) # 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 """ dummyNode = ListNode(None) dummyNode.next = head p, q = dummyNode, dummyNode for i in xrange(n+1): q = q.next while q: p = p.next q = q.next p.next = p.next.next return dummyNode.next