Reverse Nodes in k-Group[leetcode]遞迴和非遞迴的解法

標籤：leetcode

題目不難，但是容易出錯，需要考慮各種邊界情況

非遞迴代碼如下：

ListNode *reverseKGroup(ListNode *head, int k) {        if (head == NULL || k <= 1) return head;        ListNode * start = NULL, * end = NULL, *newHead = NULL, *preEnd = NULL;        while (true)        {            preEnd = end;            if (!check(head, k))            {                if (preEnd) preEnd->next = head;                if (!newHead) newHead = head;                return newHead;            }            reverse(head, k, &start, &end, &head);            if (!newHead) newHead = start;            if (preEnd) preEnd->next = start;        }        end->next = NULL;        return newHead;    }        bool check(ListNode* head, int k)    {        for (int i = 1; i <= k; i++, head = head->next)        {            if (head == NULL)            {                return false;            }        }        return true;    }        void reverse(ListNode *head, int k, ListNode** newHead, ListNode ** newTail, ListNode ** nextHead)    {        ListNode * pre, * next = head->next, * cur = head;        * newTail = cur;        for (int i = 2; i <= k; i++)        {            pre = cur;            cur = next;            next = cur->next;            cur->next = pre;        }        * newHead = cur;        * nextHead = next;    }

如果用遞迴寫會簡單些：

ListNode *reverseKGroup(ListNode *head, int k) {        if (k <= 1 || !check(head, k)) return head;        ListNode* start, *end;        reverse(head, k, &start, &end, &head);        end->next = reverseKGroup(head, k);        return start;    }        bool check(ListNode* head, int k)    {        for (int i = 1; i <= k; i++, head = head->next)            if (head == NULL)                return false;        return true;    }        void reverse(ListNode *head, int k, ListNode** newHead, ListNode ** newTail, ListNode ** nextHead)    {        ListNode * pre, * next = head->next, * cur = head;        * newTail = cur;        for (int i = 2; i <= k; i++)        {            pre = cur;            cur = next;            next = cur->next;            cur->next = pre;        }        * newHead = cur;        * nextHead = next;    }

Reverse Nodes in k-Group[leetcode]遞迴和非遞迴的解法