/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *getMid(ListNode* head){
        if(head==NULL || head->next==NULL)return head;
        ListNode*prev=NULL;
        ListNode*p1=head;
        ListNode*p2=head;
        while(p2){
            prev=p1;
            p1=p1->next;
            p2=p2->next;
            if(p2)p2=p2->next;
        }
        prev->next=NULL;
        return p1;
    }
    ListNode *sortList(ListNode *head) {
        // nlogn的复杂度，用归并排序求解
        if(head==NULL || head->next==NULL)return head;
        //原列表平分为两个列表
        ListNode*head1=head;
        ListNode*head2=getMid(head);
        //分别对两个列表进行排序
        ListNode*p1=sortList(head1);
        ListNode*p2=sortList(head2);
        ListNode*newhead=NULL;
        ListNode*p=NULL;
        //归并
        while(p1&&p2){
            if(p1->val<p2->val){
                if(p==NULL)newhead=p1;
                else p->next=p1;
                p=p1;
                p1=p1->next;
            }
            else{
                if(p==NULL) newhead=p2;
                else p->next=p2;
                p=p2;
                p2=p2->next;
            }
        }
        if(p1) p->next=p1;
        if(p2) p->next=p2;
        
        return newhead;
    }
};