C++ || Straight forward code || Simplest Approach
\nclass Solution {\npublic:\n    ListNode* mergeInBetween(ListNode* list1, int a, int b, ListNode* list2) {\n        ListNode *ptr=list1;\n         ListNode *pt
```\nclass Solution {\npublic:\n    ListNode* mergeInBetween(ListNode* list1, int a, int b, ListNode* list2) {\n        ListNode *ptr=list1;\n         ListNode *ptr2=list1;\n        while(a>1)\n        {\n            a--;\n            ptr=ptr->next;\n        }\n        while(b)\n        {\n            ptr2=ptr2->next;\n            b--;\n        }\n        ptr->next=list2;\n        while(ptr->next)\n        {\n            ptr=ptr->next;\n        }\n         ptr->next=ptr2->next;\n        return list1;\n    }\n};\n```\nPlz upvote, it means a lot.
['Linked List', 'C', 'C++']