C++ easy solution || C++
Intuition\nTo solve this problem, we need to find the ath and (b+1)th nodes in the list1 and then adjust the pointers to insert list2 between them.\n# Approach\
# Intuition\nTo solve this problem, we need to find the ath and (b+1)th nodes in the list1 and then adjust the pointers to insert list2 between them.\n# Approach\n1) Create a dummy node to simplify handling edge cases.\n2) Traverse list1 to find the ath node and the (b+1)th node.\n3) Traverse list2 to find its tail.\n4) Adjust pointers to insert list2 between the ath and (b+1)th nodes.\n5) Return the modified list1.\n# Complexity\n- Time complexity:\nO(n + m), where n is the size of list1 and m is the size of list2.\n- Space complexity:\nO(1), as we are using constant extra space.\n# Code\n```\n/**\n * Definition for singly-linked list.\n * struct ListNode {\n *     int val;\n *     ListNode *next;\n *     ListNode() : val(0), next(nullptr) {}\n *     ListNode(int x) : val(x), next(nullptr) {}\n *     ListNode(int x, ListNode *next) : val(x), next(next) {}\n * };\n */\nclass Solution {\npublic:\n    ListNode* mergeInBetween(ListNode* list1, int a, int b, ListNode* list2) {\n        ListNode* dummy = new ListNode();\n        dummy->next = list1;\n        \n        ListNode* prevA = dummy;\n        for (int i = 0; i < a; ++i) {\n            prevA = prevA->next;\n        }\n        \n        ListNode* prevB = prevA;\n        for (int i = a; i <= b + 1; ++i) {\n            prevB = prevB->next;\n        }\n        \n        ListNode* list2Tail = list2;\n        while (list2Tail->next != nullptr) {\n            list2Tail = list2Tail->next;\n        }\n        \n        prevA->next = list2;\n        list2Tail->next = prevB;\n        \n        return dummy->next;\n    }\n};\n\n```\nUPVOTE IF YOU FINDS IT HELPFULL
['C++']