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== || head == NULL)
return head; ListNode *ans,*ansTail; bool firstTime = ,flagContinue = ;
ListNode *seghead, *tempNode = head;
while(flagContinue && tempNode)
{
seghead = tempNode = head;
flagContinue = ;
int times = ;
while(tempNode)
{
tempNode = tempNode->next;
times++;
if(times == k- && tempNode)
{
flagContinue = ;
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();
ListNode *n2 = new ListNode();
ListNode *n3 = new ListNode();
ListNode *n4 = new ListNode();
ListNode *n5 = new ListNode();
n1->next = n2;
n2->next = n3;
n3->next = n4;
n4->next = n5;
ListNode *ans;
Solution myS;
ans = myS.reverseKGroup(NULL,);
return ;
}

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