题解 | #链表中的节点每k个一组翻转#

/**
 * 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) {
        auto end = head;
        for (int i = 1; i <= k - 1; i++) {
            if (end == nullptr) return head;
            end = end->next;
        }
        if (end == nullptr) return head;
        auto nexthead = end->next;
        // 从head翻转到end
        reverseFunc(head, end);
        head->next = reverseKGroup(nexthead, k);
        return end;
    }

    ListNode* reverseFunc(ListNode* start, ListNode* end) {
        ListNode* left = nullptr;
        ListNode* curr = start;
        ListNode* right = nullptr;
        while (left != end) {
            right = curr->next;
            curr->next = left;
            left = curr;
            curr = right;
        }
        return left;
    }
};