Day14_LEETCODE_92. Reverse Linked List II.ipynb

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    "# Definition for singly-linked list.\n",
    "# class ListNode(object):\n",
    "#     def __init__(self, val=0, next=None):\n",
    "#         self.val = val\n",
    "#         self.next = next\n",
    "# Reverse Sublist in Linked List\n",
    "class Solution(object):\n",
    "    def reverseBetween(self, head, left, right):\n",
    "        \"\"\"\n",
    "        :type head: ListNode\n",
    "        :type left: int\n",
    "        :type right: int\n",
    "        :rtype: ListNode\n",
    "        \"\"\"\n",
    "        if(left==right):\n",
    "            return head\n",
    "        reversedl=[]\n",
    "        list1=[]\n",
    "        node=head\n",
    "        # Appening all values to list1\n",
    "        while node != None:\n",
    "            list1.append(node.val)\n",
    "            node=node.next\n",
    "        #Reversing from left to right\n",
    "        l=left-1\n",
    "        reversedl= list1[l:right]\n",
    "        reversedl.reverse()\n",
    "        list1[l:right]=reversedl\n",
    "\n",
    "        head = ListNode(list1[0])  # Create the first node with the first element of the list\n",
    "        current = head  # Initialize a pointer to keep track of the current node\n",
    "        \n",
    "        # Iterate over the remaining elements of the list starting from the second element\n",
    "        for i in list1:\n",
    "            node = ListNode(i)  # Create a new node for the current element\n",
    "            current.next = node  # Link the current node with the new node\n",
    "            current = node  # Move the pointer to the new node\n",
    "        \n",
    "        return head.next  # Return the head of the linked list\n"
   ]
  }
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