LeetCode之“链表”：Reverse Nodes in k-Group

　　题目链接

　　题目要求：

　　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 be changed.

　　Only constant memory is allowed.

　　For 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

　　这道题个人想法是将链表分段求逆，然后再合并。具体程序如下（有点冗余，有些代码可以合并）：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head, int k)
    {
        if(!head || !head->next)
            return head;
            
        int len = 0;
        ListNode *start = head;
        while(start)
        {
            len++;
            start = start->next;
            if(len == k)
                break;
        }
        if(len < k)
            return head;
        
        ListNode *postList = start;
        start = head;
        ListNode *prev = nullptr;
        while(start != postList)
        {
            ListNode *next = start->next;
            start->next = prev;
            prev = start;
            start = next;
        }
        
        start = prev;
        while(start->next)
            start = start->next;
        start->next = postList;
        
        return prev;
    }

    ListNode* reverseKGroup(ListNode* head, int k) {
        ListNode *dummy = new(nothrow) ListNode(INT_MIN);
        assert(dummy);
        dummy->next = head;
        ListNode *preNode = dummy;
        while(true)
        {
            preNode->next = reverseList(head, k);
            head = preNode->next;
            int cnt = 0;
            while(head)
            {
                cnt++;
                head = head->next;
                preNode = preNode->next;
                if(cnt == k)
                    break;
            }
            
            if(cnt < k)
            {
                head = dummy->next;
                delete dummy;
                dummy = nullptr;
                return head;
            }
        }
    }
};