Sort List [Leetcode]

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

Solution: divide into two parts, conquer one by one, and them merge two sorted parts.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    //Merge two sorted linked list
    ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
        if(l1==NULL) return l2;
        if(l2==NULL) return l1;
        ListNode* c1 = (l1->val<=l2->val?l1:l2); //pointing to the head with the smaller val
        ListNode* c2 = (l1->val<=l2->val?l2:l1); //pointing to the head with the larger val
        while(c1->next!=NULL && c2!=NULL){
            if(c1->val<=c2->val && c2->val<=c1->next->val){
                //insert c2 behind of c1
                ListNode* tmp = c2->next;
                c2->next = c1->next;
                c1->next = c2;
                //advance both pointers
                c1 = c1->next;
                c2 = tmp;
            }
            else
                c1 = c1->next;
        }
        if(c1->next==NULL)
            c1->next = c2;
        return (l1->val<=l2->val?l1:l2);
    }
    ListNode *sortList(ListNode *head) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        if(head==NULL || head->next==NULL)
            return head;
        //Advance slow pointer to middle node
        ListNode* slow = head;
        ListNode* fast = head->next;
        while(fast!=NULL && fast->next!=NULL){
            slow=slow->next;
            fast=fast->next->next;
        }
        ListNode* newHead = slow->next;
        slow->next=NULL;
        //Sort two parts
        head=sortList(head);
        newHead=sortList(newHead);
        //Merge two sorted parts
        return mergeTwoLists(head, newHead);
    }
};