Reverse Nodes in k-Group[leetcode]递归和非递归的解法

最新推荐文章于 2025-05-21 19:43:17 发布

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本文介绍了如何使用递归和非递归方式实现链表的分组反转，并详细讨论了边界情况的处理，提供了解决实际问题的高效算法。

题目不难，但是容易出错，需要考虑各种边界情况

非递归代码如下：

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;
    }

如果用递归写会简单些：

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;
    }