C++编程学习之LeetCode OJ

Sort List:
Sort a linked list in O(n log n) time using constant space complexity.
参考：http://blog.csdn.net/jiadebin890724/article/details/21334059
http://blog.csdn.net/worldwindjp/article/details/18986737

#include<iostream>
#define COUNT 7
using namespace std;
struct ListNode {
      int val;
      ListNode *next;
      //ListNode(int x) : val(x), next(NULL) {}
  };
class Solution {
public:
	ListNode *getMidofList(ListNode *head)
	{
		if(head==NULL||head->next==NULL)
			return head;//only one node
		ListNode *first=head,*twice=head,*Middle=NULL;
		while(twice!=NULL&&twice->next!=NULL)
		{
			twice=twice->next->next;
			Middle=first;     
			first=first->next;	
		}
		return(Middle);//return the middle node of the list
	}

	/*creat a new list for ranking the orginal list in ascending order */
	ListNode *mergeList(ListNode *former, ListNode *later)
	{
		ListNode *newList=new ListNode;
		ListNode *curr=NULL;
		curr=newList;
		while(former!=NULL&&later!=NULL)
		{
			if((former->val)>(later->val))
			{
				curr->next=later;
				later=later->next;
				
			}
			else 
			{
				curr->next=former;
				former=former->next;
			}
			curr=curr->next;
		}
		curr->next=(former!=NULL)?former:later;
		return(newList->next);
	}

    ListNode *sortList(ListNode *head) //sort the list
	{
		if(head==NULL||head->next==NULL)
		{
			return(head); //only one element
		}
			ListNode *Mid=getMidofList(head);//get the middle node 
			ListNode *Midnext=Mid->next;
			Mid->next=NULL; //divide the list into two parts
			return mergeList(sortList(head),sortList(Midnext));
    }
};
void main()
{
	Solution s;
	ListNode *head=NULL,*p=NULL,*q=NULL,*newhead=NULL;
	/*creat a list*/
	head=new ListNode;
	p=new ListNode;
	head->next=p;
	if(COUNT>=1)
		cin>>p->val;
	for(int i=2;i<=COUNT;i++)
	{
		q=new ListNode;
		cin>>q->val;
		p->next=q;
		p=q;
	}
	p->next=NULL;

	newhead=s.sortList(head->next);//pass the first node of list 
	while(newhead!=NULL)
	{
		cout<<newhead->val<<"\n";
		newhead=newhead->next;
	}
}