148. 排序链表

  有意思的题目，正常思路是插入排序，时间复杂度过高，再考虑用vector存储，然后快排，这样实现简单，但是需要额外o（n）的空间复杂度。

  考虑题目O（nlogn）时间复杂度，可以想到归并算法，这样空间复杂度就能降低，小于o（n），实际上是需要o（logn），但是已经接近于常数了。

  参考常用的归并函数，写出如下代码。主要是拆分和合并的操作有些许不同，需要注意指针的问题。

  

#include<iostream>
#include <vector>
#include <set>
#include <functional>
#include <stdlib.h>
#include <stdio.h>
#include <string>
#include <sstream>
#include <list>
#include <map>
#include <stack>
#include <algorithm>
#include<iomanip>

#define INT_MIN     (-2147483647 - 1) /* minimum (signed) int value */
#define INT_MAX       2147483647    /* maximum (signed) int value */

using namespace std;

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

 ListNode* findMid(ListNode* head)
 {
	 if (head == nullptr)
		 return nullptr;
	 ListNode* slow = head;
	 ListNode* fast = head;
	 ListNode* result = head;

	 //当只有两个元素时，slow和fast指针都没动，重叠，需要单独处理
	 if (head->next->next == nullptr)
	 {
		 result = head->next;
		 head->next = nullptr;
		 return result;
	 }
	 while (fast != nullptr && fast->next != nullptr)
	 {
		 slow = slow->next;
		 fast = fast->next->next;
	 }

	 result = slow->next;
	 slow->next = nullptr;

	 return result;
 }
 ListNode* Merge(ListNode* p1, ListNode* p2)
 {
	 ListNode* head = new ListNode(6);
	 ListNode* current = head;
	 while (p1 != nullptr && p2 != nullptr)
	 {
		 if (p1->val <= p2->val)
		 {
			 current->next = p1;
			 current = current->next;
			 p1 = p1->next;
		 }
		 else
		 {
			 current->next = p2;
			 current = current->next;
			 p2 = p2->next;
		 }
	 }
	 if (p1 != nullptr)
		 current->next = p1;
	 else if (p2 != nullptr)
		 current->next = p2;

	 current = head->next;
	 delete head;
	 head = nullptr;

	 return current;
 }
 ListNode* sortList(ListNode* head) {
	 if (head == nullptr)
		 return nullptr;

	 if (head->next == nullptr)
		 return head;
	 ListNode* mid = findMid(head);

	 head = sortList(head);
	 mid = sortList(mid);

	 head = Merge(head, mid);

	 return head;
 }
int main(int argc, char** args)
{
	ListNode* head = new ListNode(4);
	head->next = new ListNode(2);
	head->next->next = new ListNode(1);
	head->next->next->next = new ListNode(3);
	//head->next->next->next->next = new ListNode(5);
	//head->next->next->next->next->next = new ListNode(6);
	head = sortList(head);
	ListNode* p = head;
	while (p != nullptr)
	{
		cout << p->val << " ";
		p = p->next;
	}
	cout << endl;
	system("pause");
	return 0;
}