LeetCode OJ--Reverse Nodes in k-Group **

http://oj.leetcode.com/problems/reverse-nodes-in-k-group/  链表  指针  

对链表翻转的变形

#include <iostream>
using namespace std;

 struct ListNode {
     int val;
     ListNode *next;
     ListNode(int x) : val(x), next(NULL) {}
 };

 ListNode *reverse(ListNode * l2)
 {
     ListNode *n1,*n2,*before;
     n1 = l2;
     n2 = n1->next;
     before = NULL;
     while(n2)
     {
         n1->next = before;
         before = n1;
         n1 = n2;
         n2 = n2->next;
     }
     n1->next = before;
     return n1;
 }

class Solution {
public:
    ListNode *reverseKGroup(ListNode *head, int k) {
        if(k==1 || head == NULL)
            return head;

        ListNode *ans,*ansTail;
         
        bool firstTime = 1,flagContinue = 1;
        ListNode *seghead, *tempNode = head;
        while(flagContinue && tempNode)
        {
            seghead = tempNode = head;
            flagContinue = 0;
            int times = 0;
            while(tempNode)
            {
                tempNode = tempNode->next;
                times++;
                if(times == k-1 && tempNode)
                {
                    flagContinue = 1;
                    break;
                }
            }
            //break up
            if(tempNode)
            {
                head = tempNode->next;
                tempNode->next = NULL;
            }
            
            if(flagContinue)
            {
                seghead = reverse(seghead);
            }
            if(firstTime)
            {
                ans = ansTail = seghead;
                firstTime = false;
            }
            else
                ansTail->next = seghead;
            while(ansTail->next)
                ansTail = ansTail->next;
        }
        return ans;
    }
};

int main()
{
    ListNode *n1 = new ListNode(1);
    ListNode *n2 = new ListNode(2);
    ListNode *n3 = new ListNode(3);
    ListNode *n4 = new ListNode(4);
    ListNode *n5 = new ListNode(5);
    n1->next = n2;
    n2->next = n3;
    n3->next = n4;
    n4->next = n5;
    ListNode *ans;
    Solution myS;
    ans = myS.reverseKGroup(NULL,6);
    return 0;
}