[leetcode]25. Reverse Nodes in k-Group

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

If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

You may not alter the values in the nodes, only nodes itself may changed.

Only constant memory is allowed.

For example,

Given this listed 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

虽然这个题难度是hard，不过沿用k=2，即leetcode24题的思路，一次过

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     struct ListNode *next; * }; */struct ListNode* reverseKGroup(struct ListNode* head, int k) {    if((head==NULL) ||(k==1))        return head;        struct ListNode* node;    struct ListNode* node1;    int tmp[k];    node = head;    int flag = 1;   while(flag==1)    {        node1 = node;        for(int i=0; i<k; i++)        {            if(node1==NULL)                flag = 0;            else            {                tmp[i]=node1->val;                node1 = node1->next;            }        }        if(flag==1)        {            for(int i=0; i<k; i++)            {                node->val = tmp[k-1-i];                node = node->next;            }        }    }          return head;}

这个题看晚上的代码，有一个很好的思路，通过栈进行转换，别人的代码如下：

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode *reverseKGroup(ListNode *head, int k) {        ListNode* newhead = new ListNode(-1);        ListNode* tail = newhead;        ListNode* begin = head;        ListNode* end = begin;        while(true)        {            int count = k;            while(count && end != NULL)            {                end = end->next;                count --;            }            if(count == 0)            {//reverse from [begin, end)                stack<ListNode*> s;                while(begin != end)                {                    s.push(begin);                    begin = begin->next;                }                while(!s.empty())                {                    ListNode* top = s.top();                    s.pop();                    tail->next = top;                    tail = tail->next;                }            }            else            {//leave out                tail->next = begin;                break;            }        }        return newhead->next;    }};