链表的排序----链表的归并排序

 ListNode * mergeTwoLists(ListNode *l1, ListNode *l2)
    {
       ListNode *head = new ListNode(-1);
       ListNode *p = head;
       
       for(;l1&&l2; p = p->next)
       {
           if(l1->val <l2->val)
           {
               p->next = l1;
               l1 =l1->next;
           }
           else
           {
               p->next = l2;
               l2 =l2->next;
           }
           
       }
        
         p->next = l1!=nullptr?l1:l2;
         return head->next;
         
        
    }

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* sortList(ListNode* head) {
    if(head==nullptr|| head->next ==nullptr) return head;
        // get the len of the list
        ListNode * fast = head;
        ListNode * slow = head;
    
        while(fast->next!=nullptr&&fast->next->next!=nullptr)
        {
            fast = fast->next->next;
            slow = slow->next;
            
        }
        
        fast = slow;       // find the mid of the linklist
        slow = slow->next;  
        fast->next = nullptr; // cut the list to two parts;
        
        ListNode *l1 = sortList(head);
        ListNode *l2 = sortList(slow);
        
        return mergeTwoLists(l1,l2);
    }
    
    ListNode * mergeTwoLists(ListNode *l1, ListNode *l2)
    {
       ListNode *head = new ListNode(-1);
       ListNode *p = head;
       
       for(;l1&&l2; p = p->next)
       {
           if(l1->val <l2->val)
           {
               p->next = l1;
               l1 =l1->next;
           }
           else
           {
               p->next = l2;
               l2 =l2->next;
           }
           
       }
        
         p->next = l1!=nullptr?l1:l2;
         return head->next;
         
        
    }
    
};