Easy C++ Solution with Video Explanation
Video Solution\nhttps://youtu.be/fydl3gM7akI\n# Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem requires merging nodes in a
# Video Solution\nhttps://youtu.be/fydl3gM7akI\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem requires merging nodes in a linked list such that we accumulate values between zeros and create new nodes with these sums. This can be effectively done by traversing the linked list, maintaining a running sum of values until a zero is encountered, and then creating new nodes with the accumulated sum.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Dummy Node: \n- Use a dummy node to simplify edge case handling and to facilitate easy return of the result list.\n2. Pointers:\n- q is used to build the result list.\n- p is used to traverse the input list, starting from the node after the head (since the head is guaranteed to be zero).\n3. Accumulating Sum: \n- Traverse the linked list using p. Add the value of each node to sum until a zero is encountered.\n4. Creating New Nodes:\n- When a zero is encountered, it indicates the end of the current segment. Create a new node with the accumulated sum and link it to the result list. Reset sum to zero for the next segment.\n5. Return Result:\n- Finally, return dummy->next, which points to the head of the newly created list.\n# Complexity\n- Time complexity:$$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:$$O(1)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\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* mergeNodes(ListNode* head) {\n        ListNode* dummy=new ListNode(10);\n        ListNode* q=dummy;\n        ListNode* p=head->next;\n        int sum=0;\n        while(p!=NULL){\n            if(p->val!=0){\n                sum+=p->val;\n            }else{\n                ListNode* add=new ListNode(sum);\n                q->next=add;\n                q=q->next;\n                sum=0;\n            }\n            p=p->next;\n        }\n        return dummy->next;\n\n    }\n};\n```
['C++']