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
     */

    size_t GetListNodeCount(ListNode* head) {
        size_t count = 0;
        while (head != nullptr) {
          ++count;
          head = head->next;
        }
          
        return count;
    }

    ListNode* reverseKGroup(ListNode* head, int k) {
        if (head == nullptr) return head;
        if (k == 1) return head;

        auto len = GetListNodeCount(head);

        auto* pre = new ListNode(-1);
        auto* new_head = pre;
        pre->next = head;
        auto start = head;
        for (auto i = 0; i < len / k; ++i) {
          for (auto j = 0; j < k - 1; ++j) {
            auto temp = start->next;	// 本题的思路和List的第二题一样，只不过是多了一步链表长度的计算
            start->next = temp->next;	// 同样使用头插法以及添加虚拟头作为previous结点的方法
            temp->next = pre->next;
            pre->next = temp;
          }
          pre = start;
          start = start->next;
        }

        auto* ret = new_head->next;
        delete new_head; // 清理空间
      
        return ret;
    }
};