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
     */
    struct cmp{
        bool operator()(ListNode* l1,ListNode* l2){
            return l1->val>l2->val;
        }
    };
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        // write code here
        priority_queue<ListNode*,vector<ListNode*>,cmp> pq;
        for(auto list:lists)
            if(list)
                pq.emplace(list);
        auto dummy=new ListNode(0);
        ListNode* cur=dummy;
        while(!pq.empty()){
            auto list=pq.top();
            pq.pop();
            cur->next=list;
            cur=cur->next;
            list=list->next;
            if(list)
                pq.emplace(list);
        }
        return dummy->next;
    }
};