19. Remove Nth Node From End of List
- Total Accepted: 140954
- Total Submissions: 445848
- Difficulty: Easy
- Contributors: Admin
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.
Code
# 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
"""
i = 0
head_bak, n_to_last, n_plus_1_to_last = head, head, head
while head:
i += 1
if i >= n+1:
n_to_last = n_to_last.next
if i >= n+2:
n_plus_1_to_last = n_plus_1_to_last.next
head = head.next
if i == n:
return head_bak.next
else:
n_plus_1_to_last.next = n_to_last.next
return head_bak
Idea
Maintain two pointers. One is pointing to (n+1) to the last element. The other is point to n to the last element. Be careful for the corner case when there are only n elements in total.
Another idea can be seen: https://discuss.leetcode.com/topic/7031/simple-java-solution-in-one-pass