题解 | #单链表的排序#

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 *	ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
class Solution {
public:
    ListNode* mergeList(ListNode* head1,ListNode* head2)
    {
        if(!head1 || !head2) return head1 ? head1 : head2;
        ListNode* head=new ListNode(0);
        ListNode* cur=head;
        while(head1 && head2)
        {
            if(head1->val <= head2->val){
                cur->next = head1;
                head1=head1->next;
            }else{
                cur->next = head2;
                head2=head2->next;
            }
            cur=cur->next;
        }
        if(head1)
        {
            cur->next = head1;
        }
        else
        {
            cur->next = head2;
        }
        return head->next;
    }

    ListNode* sortInList(ListNode* head) {
        if(head==nullptr || !head->next) return head;
        ListNode* left=head ,*mid=head, *right=head;
        while(right && right->next)
        {
            left = mid;
            mid=mid->next;
            right = right->next->next;
        }
        left->next = nullptr;
        ListNode* left_ok = sortInList(head);
        ListNode* right_ok = sortInList(mid);
        return mergeList(left_ok,right_ok);
    }
};