✅Super Easy Fast & Slow Pointer (C++/Java/Python) Solution With Detailed Explanation✅
Intuition\nThe intuition is to first find the middle of the list using the slow and fast pointers method. Next, reverse the second half of the list starting fro
# Intuition\nThe intuition is to first find the middle of the list using the slow and fast pointers method. Next, reverse the second half of the list starting from the middle. Finally, merge the two halves by alternately linking nodes from each half. This process rearranges the list in place, requiring only a fixed number of pointers for manipulation.\n\n# Approach\n\n1. **Find the Middle of the List**: The algorithm starts by finding the middle of the linked list using the slow and fast pointer technique. This is done by advancing one pointer (`slow`) by one node and another pointer (`fast`) by two nodes in each step. When `fast` reaches the end of the list, `slow` will be at the middle. This step ensures that the list can be divided into two halves for the subsequent steps.\n\n2. **Reverse the Second Half of the List**: Once the middle of the list is found, the next step is to reverse the second half of the list starting from the `slow` pointer. This is achieved by rearranging the pointers of the nodes in the second half such that their order is reversed.\n\n3. **Merge the Two Halves**: Finally, the first half and the reversed second half are merged together to form the reordered list. This merging is done by alternating nodes from the first and the second half until the reordering criteria are met.\n\n## Complexity Analysis\n\n1. **Time Complexity**:\n   - **Finding the middle of the list**: This operation is \\(O(n)\\), where \\(n\\) is the number of nodes in the list, as it requires traversing half of the list.\n   - **Reversing the second half of the list**: This operation is also \\(O(n)\\), but since it\'s only reversing the second half, it can be considered \\(O(n/2)\\). However, for big O notation, we drop constants, so it remains \\(O(n)\\).\n   - **Merging the two halves**: The merge operation is \\(O(n)\\) because it involves traversing the entire list once to reorder the nodes.\n   - **Total Time Complexity**: The overall time complexity is \\(O(n) + O(n) + O(n) = O(n)\\), which simplifies to \\(O(n)\\) since all steps are sequential.\n\n2. **Space Complexity**:\n   - The algorithm modifies the list in place and uses a fixed number of pointer variables (`slow`, `fast`, `prev`, `next`, `firstHalf`, `secondHalf`). Thus, the space complexity is \\(O(1)\\), indicating that it requires a constant amount of additional space regardless of the input size.\n\n# Code\n```C++ []\nclass Solution {\npublic:\n    void reorderList(ListNode* head) {\n        if (head == nullptr || head->next == nullptr) return;\n\n        // Step 1: Find the middle of the list\n        ListNode* slow = head;\n        ListNode* fast = head;\n        while (fast != nullptr && fast->next != nullptr) {\n            slow = slow->next;\n            fast = fast->next->next;\n        }\n\n        // Step 2: Reverse the second half of the list\n        ListNode* prev = nullptr;\n        ListNode* next = nullptr;\n        while (slow != nullptr) {\n            next = slow->next;\n            slow->next = prev;\n            prev = slow;\n            slow = next;\n        }\n\n        // Step 3: Merge the two halves\n        ListNode* firstHalf = head;\n        ListNode* secondHalf = prev; // prev now points to the head of the reversed second half\n        while (secondHalf->next) {\n            next = firstHalf->next;\n            prev = secondHalf->next;\n\n            firstHalf->next = secondHalf;\n            secondHalf->next = next;\n            \n            firstHalf = next;\n            secondHalf = prev;\n        }\n    }\n};\n\n```\n```Java []\n\npublic class Solution {\n    public void reorderList(ListNode head) {\n        if (head == null || head.next == null) return;\n\n        // Step 1: Find the middle of the list\n        ListNode slow = head, fast = head;\n        while (fast != null && fast.next != null) {\n            slow = slow.next;\n            fast = fast.next.next;\n        }\n\n        // Step 2: Reverse the second half of the list\n        ListNode prev = null, next = null;\n        while (slow != null) {\n            next = slow.next;\n            slow.next = prev;\n            prev = slow;\n            slow = next;\n        }\n\n        // Step 3: Merge the two halves\n        ListNode firstHalf = head;\n        ListNode secondHalf = prev; // prev now points to the head of the reversed second half\n        while (secondHalf.next != null) {\n            next = firstHalf.next;\n            prev = secondHalf.next;\n            \n            firstHalf.next = secondHalf;\n            secondHalf.next = next;\n            \n            firstHalf = next;\n            secondHalf = prev;\n        }\n    }\n}\n\n```\n```Python []\n\nclass Solution:\n    def reorderList(self, head):\n        if not head or not head.next:\n            return\n\n        # Step 1: Find the middle of the list\n        slow, fast = head, head\n        while fast and fast.next:\n            slow = slow.next\n            fast = fast.next.next\n\n        # Step 2: Reverse the second half of the list\n        prev, curr = None, slow\n        while curr:\n            next_temp = curr.next\n            curr.next = prev\n            prev = curr\n            curr = next_temp\n\n        # Step 3: Merge the two halves\n        first, second = head, prev\n        while second.next:\n            # Save next pointers\n            tmp1 = first.next\n            tmp2 = second.next\n            \n            # Reorder nodes\n            first.next = second\n            second.next = tmp1\n            \n            # Move pointers forward\n            first = tmp1\n            second = tmp2\n\n```
['Linked List', 'Two Pointers', 'C++', 'Java', 'Python3']