C# Solution for Merge In Between Linked Lists Problem
Intuition\n Describe your first thoughts on how to solve this problem. \nThe intuition behind the solution is to traverse through the linked list until reaching
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind the solution is to traverse through the linked list until reaching the node before the range to be replaced (a-1th node). Then, skip the nodes within the range (ath to bth node) and connect the a-1th node to the start of list2. After that, traverse through list2 to find its end and connect it to the node after the bth node.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1.\tTraverse through the linked list to reach the (a-1)th node and store it in prev.\n2.\tTraverse through the linked list again to reach the bth node and store it in curr.\n3.\tConnect the next of prev to list2.\n4.\tTraverse through list2 to find its end.\n5.\tConnect the next of the last node of list2 to curr.next.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\u2022\tTraversing through the linked list twice: O(n), where n is the number of nodes in list1.\n\u2022\tTraversing through list2 to find its end: O(list2.length)\nOverall, the time complexity is O(n + list2.length).\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThe space complexity is O(1) because we are using a constant amount of extra space regardless of the input size.\n\n# Code\n```\n/**\n * Definition for singly-linked list.\n * public class ListNode {\n *     public int val;\n *     public ListNode next;\n *     public ListNode(int val=0, ListNode next=null) {\n *         this.val = val;\n *         this.next = next;\n *     }\n * }\n */\npublic class Solution {\n    public ListNode MergeInBetween(ListNode list1, int a, int b, ListNode list2) {\n        ListNode dummy = new ListNode(0);\n        dummy.next = list1;\n        ListNode prev = dummy;\n        \n        // Move prev to the (a-1)th node\n        for (int i = 0; i < a; i++) {\n            prev = prev.next;\n        }\n        \n        // Move curr to the bth node\n        ListNode curr = prev;\n        for (int i = a; i <= b; i++) {\n            curr = curr.next;\n        }\n        \n        // Connect the (a-1)th node to list2\n        prev.next = list2;\n        \n        // Find the end of list2\n        while (list2.next != null) {\n            list2 = list2.next;\n        }\n        \n        // Connect the end of list2 to the node after the bth node\n        list2.next = curr.next;\n        \n        return dummy.next;\n    }\n}\n```
['C#']