class Solution { public: ListNode* reverseKGroup(ListNode* head, int k) { if (head == nullptr || k == 1) return head; ListNode dummy(0); dummy.next = head; ListNode* prev = &dummy; // 指向待反转组的前一个节点 while (true) { // 检查剩余节点是否够k个 ListNode* tail = prev; for (int i = 0; i < k; i++) { tail = tail->next; if (tail == nullptr) return dummy.next; // 不足k个，直接返回 } // 反转当前组的k个节点 ListNode* curr = prev->next; ListNode* next = curr->next; ListNode* groupTail = curr; // 当前组的第一个节点会成为反转后的最后一个节点 for (int i = 1; i < k; i++) { ListNode* temp = next->next; next->next = curr; curr = next; next = temp; } // 连接反转后的组 prev->next = curr; groupTail->next = next; // 移动prev到下一组的前一个节点 prev = groupTail; } return dummy.next; } };
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class Solution { public: ListNode* reverseKGroup(ListNode* head, int k) { // 检查是否有k个节点 ListNode* curr = head; int count = 0; while (curr != nullptr && count < k) { curr = curr->next; count++; } // 不足k个，直接返回原链表 if (count < k) return head; // 反转前k个节点 ListNode* prev = nullptr; ListNode* next = nullptr; curr = head; for (int i = 0; i < k; i++) { next = curr->next; curr->next = prev; prev = curr; curr = next; } // 递归反转剩余部分 head->next = reverseKGroup(curr, k); return prev; } };
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