Leetcode: Remove Nth Node From End of List

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
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