题解 | #牛群的重新分组#

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 *	ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param head ListNode类 
     * @param k int整型 
     * @return ListNode类
     */
    ListNode* reverseKGroup(ListNode* head, int k) 
    {
        // write code here
        vector<int> data;
        while(head != NULL)
        {
            data.push_back(head->val);
            head = head->next;
        }
        int n = data.size();
        for (int i = 0; i < n; i += k) 
        {
            int left = i;
            int right = min(i + k - 1, n - 1); // Ensure not to go beyond array bounds
            if (right - left + 1 == k) 
            { // Check if there are enough elements to reverse
                while (left < right) 
                {
                    swap(data[left], data[right]);
                    left++;
                    right--;
                }
            }
        }
        ListNode* prev = new ListNode(0);
        ListNode* result = prev;
        for (auto it : data)
        {
            ListNode* cur = new ListNode(it);
            prev->next = cur;
            prev = cur;
        }
        return result->next;
    }
};