题解 | #合并k个已排序的链表#

/**
 * 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 lists ListNode类vector 
     * @return ListNode类
     */
    ListNode* mergeKLists(vector<ListNode*>& lists) 
    {
        vector<int>record;
        for(int i=0;i<lists.size();i++)
        {
            if(lists[i]==nullptr)
            {
                record.push_back(i);
            }
        }
        if(record.size()==lists.size())
        {
            return nullptr;
        }
        else
        {
            sort(record.begin(), record.end(), greater<>());
            for(const auto &it :record)
            {
                delete lists[it];
                lists.erase(lists.begin() + it);
            }
        }


        int val=lists[0]->val;
        int num=0;
        for(int i=0;i<lists.size();i++)
        {
            if(lists[i]->val<val)
            {
                val=lists[i]->val;
                num=i;
            }
        }
        auto *q=lists[num];
        lists[num]=lists[num]->next;
        delete q;


        auto *p=new ListNode(val);
        p->next=mergeKLists(lists);
        return p;
    }
};