Exchange Two Symmetric List Values

### Problem Given linked-list `head` and one-based position `k`, exchange the values stored in the kth node from the start and the kth node from the end. Return the existing list head after this value mutation. List links need not change. ```cpp class Solution { public: ListNode* swapNodes(ListNode* head, int k); }; ``` ### Examples **Example 1:** ```text List values: [2,4,6,8], k = 1 Output list values: [8,4,6,2] Explanation: The first and last node values exchange. ``` **Example 2:** ```text List values: [3,7,9,1,5], k = 3 Output list values: [3,7,9,1,5] Explanation: Both selected positions refer to the middle node. ``` ### Constraints - The list has n nodes, with `1 <= k <= n <= 10^5`. - Each node value is in `[0,100]`. ### Clarifying Questions The original API receives a ListNode pointer and returns a ListNode pointer. Swap stored values, as the source states, rather than returning an array or rebuilding unrelated nodes. ```hint Inspect the structure Which existing nodes are selected by the two one-based positions, and which part of each node may change? ``` ### Reasoning checkpoints - Locate both one-based positions without an off-by-one error. - Mutate their values while preserving links and head identity. ### Optional extensions How would swapping the node objects themselves complicate adjacent-node and head cases?