{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<h1>876. Middle of the Linked List</h1>\n",
    "<hr>\n",
    "\n",
    "<!--Copy Paste Leetcode statement between-->\n",
    "<p>Given a non-empty, singly&nbsp;linked list with head node <code>head</code>, return&nbsp;a&nbsp;middle node of linked list.</p>\n",
    "\n",
    "<p>If there are two middle nodes, return the second middle node.</p>\n",
    "\n",
    "<p>&nbsp;</p>\n",
    "\n",
    "<div>\n",
    "<p><strong>Example 1:</strong></p>\n",
    "\n",
    "<pre><strong>Input: </strong><span id=\"example-input-1-1\">[1,2,3,4,5]</span>\n",
    "<strong>Output: </strong>Node 3 from this list (Serialization: <span id=\"example-output-1\">[3,4,5]</span>)\n",
    "The returned node has value 3.  (The judge's serialization of this node is [3,4,5]).\n",
    "Note that we returned a ListNode object ans, such that:\n",
    "ans.val = 3, ans.next.val = 4, ans.next.next.val = 5, and ans.next.next.next = NULL.\n",
    "</pre>\n",
    "\n",
    "<div>\n",
    "<p><strong>Example 2:</strong></p>\n",
    "\n",
    "<pre><strong>Input: </strong><span id=\"example-input-2-1\">[1,2,3,4,5,6]</span>\n",
    "<strong>Output: </strong>Node 4 from this list (Serialization: <span id=\"example-output-2\">[4,5,6]</span>)\n",
    "Since the list has two middle nodes with values 3 and 4, we return the second one.\n",
    "</pre>\n",
    "\n",
    "<p>&nbsp;</p>\n",
    "\n",
    "<p><strong>Note:</strong></p>\n",
    "\n",
    "<ul>\n",
    "\t<li>The number of nodes in the given list will be between <code>1</code>&nbsp;and <code>100</code>.</li>\n",
    "</ul>\n",
    "</div>\n",
    "</div>\n",
    "<!--Copy Paste Leetcode statement between-->\n",
    "\n",
    "<p>&nbsp;</p>\n",
    "<a href=\"https://leetcode.com/problems/middle-of-the-linked-list/\">Source</a> \n",
    "<hr>"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [],
   "source": [
    "# Definition for singly-linked list.\n",
    "class ListNode:\n",
    "    def __init__(self, val=0, next=None):\n",
    "        self.val = val\n",
    "        self.next = next"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<h4>Code</h4>"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [],
   "source": [
    "def middle_node(head):\n",
    "    \"\"\"Naive solution: Counting the number of nodes\"\"\"\n",
    "    node = head\n",
    "\n",
    "    counter = 0\n",
    "    while node.next:\n",
    "        counter += 1\n",
    "        node = node.next\n",
    "\n",
    "    counter = (counter+1) // 2   # counter+1 to account for the last node\n",
    "    node = head\n",
    "    while counter > 0:\n",
    "        counter -= 1\n",
    "        node = node.next\n",
    "    return node"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [],
   "source": [
    "def middle_node(head):\n",
    "    \"\"\"Using 2 pointers.\"\"\"\n",
    "    if head is None:\n",
    "        return head\n",
    "    slow, fast = head, head\n",
    "    while fast and fast.next:\n",
    "        slow = slow.next\n",
    "        fast = fast.next.next\n",
    "    return slow"
   ]
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "Python 3",
   "language": "python",
   "name": "python3"
  },
  "language_info": {
   "codemirror_mode": {
    "name": "ipython",
    "version": 3
   },
   "file_extension": ".py",
   "mimetype": "text/x-python",
   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython3",
   "version": "3.8.2"
  }
 },
 "nbformat": 4,
 "nbformat_minor": 1
}