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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
Linked List
#include<iostream>
using namespace std;
struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};
ListNode* reverseKGroup(ListNode* head, int k) {
if(k<=1||head==NULL)
return head;
ListNode *ptr0=head;
ListNode *ptr1=head;
ListNode *head1=NULL;
ListNode *ptr2=NULL;
int N=k-1;
int flag=0;
while(N--)
{
ptr1=ptr1->next;
if(ptr1==NULL)
{
flag=1;
break;
}
}
if(flag==1)
return ptr0;
ptr2=ptr1->next;
ListNode *ptr3=ptr0->next;
if(ptr3==ptr1)
{
ptr1->next=ptr0;
ptr0->next=reverseKGroup(ptr2,k);
return ptr1;
}
ListNode *ptr4=NULL;
ptr0->next=reverseKGroup(ptr2,k);
while(1)
{
ptr4=ptr3->next;
ptr3->next=ptr0;
if(ptr4==ptr1)
break;
else
{
ptr0=ptr3;
ptr3=ptr4;
}
}
ptr1->next=ptr3;
return ptr1;
}
int main()
{
}
leetcode_25题——Reverse Nodes in k-Group （链表）
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原文地址：http://www.cnblogs.com/yanliang12138/p/4592114.html
