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* reverseBetween(ListNode* head, int n) {
        // write code here
        auto* dummy = new ListNode(-1);
        dummy->next = head;
        auto pre = dummy;
        head = pre->next;
        for (int i = 1; i < n; ++i) {
            auto next = head->next;
            head->next = next->next;
            next->next = pre->next;
            pre->next = next;
        }
        return dummy->next;
    }
    ListNode* reverseKGroup(ListNode* head, int k) {
        // write code here
        int length = 0;
        auto dummy = new ListNode(-1);//创建一个虚拟节点
        auto dummy1 = dummy;//创建每一组翻转的虚拟节点
        dummy->next = head;//让虚拟节点的next指向头节点，最后返回直接返回dummy->next
        while (head != nullptr) {//先遍历链表，确定链表的长度
            ++length;
            head = head->next;
        }
        auto times = static_cast<int>(length / k);//确定需要反转的组数
        while (times--) {
            auto pre = dummy1;//相当于将每一组链表独立出来，每一组的解法同BM2
            head = pre->next;
            for (int i = 1; i < k; ++i) {
                auto next = head->next;
                head->next = next->next;
                next->next = pre->next;
                pre->next = next;
            }
            dummy1 = head;//翻转以后，头节点必定是这一组节点的尾节点，同时也必定是下一组头节点的前驱节点
        }
        return dummy->next;

    }
};