[LeetCode] Challenge log 19

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19. Remove Nth Node From End of List

The trick is to create two pointers. One is n steps faster.

Given a linked list, remove the n-th node from the end of list and return its head.

Example:

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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?

Soulution:
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# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        if head is None: 
            return None
        
        # slow = fast - n
        fast = head
        slow = head
        while n > 0:
            fast = fast.next
            n -= 1
            
        if fast is None: # if head is to be delete
            return head.next
        else: # otherwise
            while fast.next:
                fast = fast.next
                slow = slow.next
            slow.next = slow.next.next
            return head
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