Leetcode.25. Reverse Nodes in k-Group

题意：链表翻转，每k个翻一次，最后不足k个的不翻

题解：没啥难点，就是要对逻辑有一点要求；当然你也可以先存到数组里，然后用数学方法计算下标进行swap，但是这脱离了这道题的本意，代码不放了

class Solution {
public:
    void reverse(vector<ListNode*>& lists) {
        for (auto i = 0; i < lists.size() - 1; i++) {
            ListNode *st = lists[i], *ed = lists[i + 1];
            ListNode *cur = st, *nex = st->next, *tmp;
            while (nex != ed) {
                tmp = nex->next;
                nex->next = cur;
                cur = nex;
                nex = tmp;
            }
            st = cur;
            lists[i] = st;
        }
        for (auto i = 0; i < lists.size() - 1; i++) {
            ListNode *st = lists[i], *ed = lists[i + 1];
            while (st->next->next != st) st = st->next;
            st->next->next = ed;
        }
        return;
    }
    
    ListNode* reverseKGroup(ListNode* head, int k) {
        if (head == NULL || k <= 1)
            return head;
        vector<ListNode*> lists;
        ListNode* p = head;
        int cnt = 0;
        while (p != NULL)
        {
            for (auto i = 0; i < k && p != NULL; i++) {
                if (i == 0)
                    lists.push_back(p);
                cnt++;
                p = p->next;
            }
        }
        if (cnt % k == 0)
            lists.push_back(NULL);
        reverse(lists);
        return lists[0];
    }
};