Day15_LEETCODE_19. Remove Nth Node From End of List.ipynb

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    "# Definition for singly-linked list.\n",
    "# class ListNode:\n",
    "#     def __init__(self, val=0, next=None):\n",
    "#         self.val = val\n",
    "#         self.next = next\n",
    "class Solution:\n",
    "    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:\n",
    "        listlen=0 ; currentn=head        \n",
    "        # Checking length of linkedlist\n",
    "        while currentn:\n",
    "            listlen+=1\n",
    "            currentn=currentn.next      \n",
    "        current=head ; count=1 ; prevnode=current ; nextnode= current       \n",
    "        pos=listlen-n + 1 # Position of nth node from the end         \n",
    "               \n",
    "        if listlen ==1 and n==1:\n",
    "            return None\n",
    "        \n",
    "        elif listlen>1:\n",
    "            #Traversing the list till nth node from the end\n",
    "            while count <= pos:                \n",
    "                if count==pos-1: # To get the node before the nth node from the end\n",
    "                    prevnode=current\n",
    "                    current=current.next\n",
    "                    count+=1                               \n",
    "                elif count == pos: # When at nth node from the end\n",
    "                    if current== head: # If first node is to be removed\n",
    "                        head=current.next\n",
    "                    elif n ==listlen: # If last node is to be removed\n",
    "                        prevnode.next = None \n",
    "                    elif n!= listlen: # If node to be removed is not the last node\n",
    "                        nextnode= current.next                      \n",
    "                        current.next=None # Making node at nth position point to null\n",
    "                        prevnode.next = nextnode # Making node before nth node point to node after nth node\n",
    "                    count+=1                \n",
    "                else :\n",
    "                    current=current.next\n",
    "                    count+=1\n",
    "            return head\n",
    "\n"
   ]
  }
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