[LeetCode] 19. Remove Nth Node From End of List

Problem : https://leetcode.com/problems/remove-nth-node-from-end-of-list/

Use 2 pointers to iterate the linked list. The second pointer is n steps behind the first pointer.

Time Complexity :  O ( N )

Space Complexity : O ( 1 )

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        
        # use dummy head node to handle the case that needs to remove the first node.
        dummy = ListNode(0)
        dummy.next = head
        
        p1, p2 = dummy, dummy
        
        while p2 and n >= 0:
            p2 = p2.next
            n -= 1
            
        while p2 and p1:
            p2 = p2.next
            p1 = p1.next
            
        if p1 and p1.next:
            p1.next = p1.next.next
    
        
        return dummy.next