首页 > 其他 > 详细

Reverse Nodes in k-Group--LeetCode

时间:2015-03-31 18:01:43      阅读:255      评论:0      收藏:0      [点我收藏+]

题目:

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

思路:先反转链表,然后再连接

#include <iostream>
#include <vector>
using namespace std;
/*
按照规定 做K反转 
*/
typedef struct list_node List;
struct list_node
{
	int value;
	struct list_node* next;
};

void Init_List(List*& head,int* array,int n)  
{  
    head = NULL;  
    List* tmp;  
    List* record;  
    for(int i=1;i<=n;i++)  
    {  
        tmp = new List;  
        tmp->next = NULL;  
        tmp->value = array[i-1];  
        if(head == NULL)  
        {  
            head = tmp;  
            record = head;  
        }  
        else  
        {  
            record->next = tmp;  
            record = tmp;  
        }  
    }  
}  
void print_list(List* list)  
{  
    List* tmp=list;  
    while(tmp != NULL)  
    {  
        cout<<tmp->value<<endl;  
        tmp = tmp->next;   
    }  
}  


/*反转链表List 是的新的头部为head 新的尾部为tail*/
void Reverse_list(List*& list,List*& head,List*& tail)
{
	if(list == NULL || list->next == NULL)
		return ;
	head = list;
	tail = list;
	List* cur=NULL;
	List* next;
	while(head !=NULL)
	{
		next = head->next;
		head->next=cur ;
		cur = head;
		head = next;
	}
	list =cur;
	head = cur;
}

/*做K个节点的反转*/
void Reverse_k(List*& list,int k)
{
	int num = 1;
	int flag =1;
	if(list == NULL || list->next == NULL || k<=0)
		return;
	List* head,*tail,*next,*pre;
	head = list;
	tail = list;
	while(tail != NULL && tail->next != NULL)
	{
		tail = tail->next;
		num++;
		if(num == k)
		{
			if(tail != NULL)
			{
				next = tail->next;
				tail->next = NULL;		
			}		
			else
				next = NULL;
			print_list(head);
			cout<<"============"<<endl;
			Reverse_list(head,head,tail);
			
			if(flag)
			{
				list = head;
				flag =0;
				pre = tail;
			}
			else //第二次以后的反转 
			{
				pre->next = head;
				pre = tail; 
			}
			//tail->next = next;
			head = next;
			tail = next;
			num =1;
		}
	}
	pre->next = head;
}
   
int main() 
{
	int array[]={1,2,3,4,5,6,7,8,9,10,11,12};
	List* list,*head,*tail;
	Init_List(list,array,sizeof(array)/sizeof(int));
	Reverse_k(list,3);	
	print_list(list); 
	return 0;
}


Reverse Nodes in k-Group--LeetCode

原文:http://blog.csdn.net/yusiguyuan/article/details/44782415

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!