25. Reverse Nodes in k-Group

description:

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.
Note:

Example:

Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

my answer:

感恩

画图，没有难点，就是来回挪，一个循环找组，就是找到k个数看有没有，另一个函数在组内翻转，翻转的原理就是一个一个的把节点挪到第一个的位置上，其他的后挪。

大佬的answer：

class Solution {
public:
    ListNode* reverseKGroup(ListNode* head, int k) {
        if (!head || k == 1) return head;//这里是返回原列表不是返回NULL
        ListNode *dummy = new ListNode(-1), *pre = dummy, *cur = head;//这里cur = head,而不是cur = pre->next，因为此时pre的下一个是啥不知道，之后才会定义
        dummy->next = head;
        for (int i = 1; cur; ++i) {//这里是从1开始，条件是cur有意义，这里的i并没有传统意义上的计数，而是一个外部的技术，迭代的不是i，是cur
            if (i % k == 0) {
                pre = reverseOneGroup(pre, cur->next);
                cur = pre->next;
            } else {
                cur = cur->next;
            }
        }
        return dummy->next;
    }
    ListNode* reverseOneGroup(ListNode* pre, ListNode* next) {
        ListNode *last = pre->next, *cur = last->next;
        while(cur != next) {
            last->next = cur->next;
            cur->next = pre->next;
            pre->next = cur;
            cur = last->next;
        }
        return last;
    }
};

relative point get√：

hint :