19. Remove Nth Node From End of List leetcode Python 2016 new Season

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.

Show Tags
Have you met this question in a real interview? 
Yes
 
No

Discuss


this problem can be done by doing with a fast pointer and a slow pointer.

O(n) time 

# 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
        """
        dummyHead = ListNode(None)
        dummyHead.next = head
        fast = slow = dummyHead
        for _ in range(n):
            if fast:
                fast = fast.next
        if n == 0 or not fast:
            return dummyHead.next
        while fast.next:
            fast = fast.next
            slow = slow.next
        if slow.next:
            slow.next = slow.next.next
        return dummyHead.next
        


你可能感兴趣的:(19. Remove Nth Node From End of List leetcode Python 2016 new Season)