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:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param head ListNode类 the head node
     * @return ListNode类
     */
     //合并两个链表
     ListNode* marge(ListNode* Head1, ListNode* Head2)
     {
        if(Head1 == nullptr)
        {
            return Head2;
        }
        if(Head2 == nullptr)
        {
            return Head1;
        }

        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;
        }
        if(Head2)
        {
            cur->next = Head2;
        }
        return head->next;
     }
    ListNode* sortInList(ListNode* head) {
        if(head == nullptr || head->next == nullptr)
            return head;
        
        ListNode* left = head;
        ListNode* mid = head->next;
        ListNode* right = head->next->next;
        while(right != nullptr && right->next != nullptr)
        {
            left = left->next;
            mid = mid->next;
            right = right->next->next;
        }

        left->next = nullptr;

        return marge(sortInList(head), sortInList(mid));
        
    }
};