题解 | #链表中的节点每k个一组翻转#

链表中的节点每k个一组翻转

https://www.nowcoder.com/practice/b49c3dc907814e9bbfa8437c251b028e

/**
 * struct ListNode {
 *  int val;
 *  struct ListNode *next;
 * };
 */

class Solution {
  public:
    /**
     *
     * @param head ListNode类
     * @param k int整型
     * @return ListNode类
     */
    ListNode* reverseKGroup(ListNode* head, int k) {
        // write code here
        ListNode* dummyHead = new ListNode(0);
        dummyHead->next = head;
        ListNode* p = dummyHead, * q = dummyHead;
        int step;
        while (true) {
            q = p;
            for (step = k; q->next != nullptr && step != 0; step--) q = q->next;
            if (step != 0) return dummyHead->next;
            ListNode* pre = q->next, * cur = p->next;
            while (pre != q) {
                ListNode* temp = cur->next;
                cur->next = pre;
                pre = cur;
                cur = temp;
            }
            p->next = q;
            for (step = k; p->next != nullptr && step != 0; step--) p = p->next;
        }
    }
};

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