题解 | #链表中的节点每k个一组翻转#

/**
 * 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
        auto dummy = new ListNode(-1), p = dummy;
        dummy->next = head;
        while (true) {
            auto q = p;
            for (int i = 0; i < k && q; i++) q = q->next;
            if (!q) break;
            auto a = p->next, b = a->next;
            for (int i = 0; i < k - 1; i++) {
                auto temp = b->next;
                b->next = a;
                a = b;
                b = temp;
            }
            auto c = p->next;
            c->next = b;
            p->next = a;
            p = c;
        }
        return dummy->next;
    }
};

/**
 * 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
        auto dummy = new ListNode(-1);
        dummy->next = head;
        auto pre = dummy;
        while (true) {
            auto p = pre;
            for (int i = 0; i < k; i++) {
                if (!p->next) return dummy->next;
                p = p->next;
            }
            auto cur = pre->next;
            for (int i = 0; i < k - 1; i++) {
                auto temp = cur->next;
                cur->next = temp->next;
                temp->next = pre->next;
                pre->next = temp;
            }
            pre = cur;
        }
        return dummy->next;
    }
};

• 思路：分组反转
• 1、设置空节点
• 2、pre指针指向反转组的前一个节点
• 3、首先判断反转组内节点数是否达到k个，不满足k个则直接返回
• 4、反转组内节点，然后将调整指针
• 时间复杂度：O(n)
• 空间复杂度：O(1)