题解 | #单链表的排序# 亲自写一遍，再看看吧

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param head ListNode类 the head node
     * @return ListNode类
     */
    ListNode* sortInList(ListNode* head) {
        if (!head || !head->next) {
            return head;
        }
        ListNode* mid = getMid(head);
        ListNode *left = sortInList(head);
        ListNode *right = sortInList(mid);
        return Merge(left, right);        
    }
    
    ListNode* getMid(ListNode* head) {
        if(!head) {
            return head;
        }
        ListNode* fast = head, *slow = head;
        ListNode* prev = slow;
        while(fast && fast->next) {
            prev = slow;
            slow = slow->next;
            fast = fast->next->next;
        }
        prev->next = nullptr; // 切断mid和之前的联系，mid属于后面
        return slow;
    }
    
    ListNode *Merge(ListNode *l1, ListNode *l2) {
        if (!l1) {
            l2;
        }
        if (!l2) {
            l1;
        }
        ListNode *head = new ListNode(-1);
        ListNode *ret = head;
        while(l1 && l2) {
            if (l1->val < l2->val) {
                ret->next = new ListNode(l1->val);
                l1 = l1->next;
            } else {
                ret->next = new ListNode(l2->val);
                l2 = l2->next;
            }
            ret = ret->next;
        }
        if (l1) {
            ret->next = l1;
        }
        if (l2) {
            ret->next = l2;
        }
        return head->next;
    }
    

};