LeetCode148：Sort List

题目：

Sort a linked list in O(n log n) time using constant space complexity.

解题思路：

根据题目要求，可知只能用归并排序，其他排序算法要么时间复杂度不满足，要么空间复杂度不满足

实现代码：

#include <iostream>



using namespace std;

/*

Sort a linked list in O(n log n) time using constant space complexity.

*/

/**

 * Definition for singly-linked list.

 * struct ListNode {

 *     int val;

 *     ListNode *next;

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

 * };

 */

struct ListNode

{

    int val;

    ListNode *next;

    ListNode(int x):val(x), next(NULL)

    {



    }



    

};



void addNode(ListNode* &head, int val)

{

    ListNode *newNode = new ListNode(val);

    if(head == NULL)

    {

        head = newNode;

    }

    else

    {

        newNode->next = head;

        head = newNode;

    }

}



void PrintList(ListNode *root)

{

    ListNode *head = root;

    while(head != NULL)

    {

        cout<<head->val<<"\t";

        head = head->next;

    }

    cout<<endl;

}



class Solution {

public:

    ListNode *sortList(ListNode *head) {

           if(head == NULL || head->next == NULL)

           return head;

        ListNode *quick = head;

        ListNode *slow = head;

        while(quick->next && quick->next->next)//通过两个指针，一个走两步、一个走一步，获得链表的中间节点 

        {

            slow = slow->next;

            quick = quick->next->next;            

        }

        quick = slow;

        slow = slow->next;

        quick->next = NULL;//将链表的前半段进行截断 

        ListNode *head1 = sortList(head);

        ListNode *head2 = sortList(slow);

        return merge(head1, head2);

    }

    

    //归并两个有序链表 

    ListNode *merge(ListNode *head1, ListNode *head2)

    {

        if(head1 == NULL)

            return head2;

        if(head2 == NULL)

            return head1;

        ListNode *newHead = NULL;

        if(head1->val < head2->val)

        {

            newHead = head1;

            head1 = head1->next;

            

        }

        else

        {

            newHead = head2;

            head2 = head2->next;    

        }

        ListNode *p = newHead;

        while(head1 && head2)

        {

            if(head1->val < head2->val)

            {

                p->next = head1;

                head1 = head1->next;

            }

            else

            {

                p->next = head2;

                head2 = head2->next;

            }

            p = p->next;

        }

        if(head1)

            p->next = head1;

        if(head2)

            p->next = head2;

        return newHead;

    }

};



int main(void)

{

    ListNode *head = new ListNode(5);

    addNode(head, 3);

    addNode(head, 10);

    addNode(head, 15);

    PrintList(head);

    

    Solution solution;

    head = solution.sortList(head);

    PrintList(head);

    

    return 0;

}