LinkedList Easy Solution
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n/**\n * Definition for singly-linked list.\n * public class ListNode {\n *     int val;\n *     ListNode next;\n *     ListNode() {}\n *     ListNode(int val) { this.val = val; }\n *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }\n * }\n */\nclass Solution {\n    public ListNode mergeInBetween(ListNode list1, int a, int b, ListNode list2) {\n         ListNode temp1 =list1;\n          ListNode temp2 =list1;\n         for(int i=0;i<a-1;i++){\n            temp1 = temp1.next;\n            }\n        for(int i=0;i<b+1;i++){\n            temp2 = temp2.next;\n            }\n            //System.out.println(temp1.val+" "+temp2.val);\n            // I put pointer temp1 and temp2;\n            temp1.next=list2;\n            ListNode pointer  = list2;\n\n            while(pointer.next != null){\n                pointer  = pointer.next;\n            } \n            pointer.next  = temp2;\n           return list1;\n    }\n}\n```
['Java']