题解 | #合并k个已排序的链表#

class Solution {
public:
    ListNode *mergeKLists(vector<ListNode *> &lists) {
        if(lists.empty()) return nullptr;
        struct cmp_custom {//比较器
            bool operator()(ListNode* &x, ListNode* &y) {
                return (x->val) > (y->val);//递减
            }
        };
        priority_queue<ListNode*,vector<ListNode*>,cmp_custom> p;
        for(int i=0;i<lists.size();++i){
            if(lists[i])
                p.push(lists[i]);
        }
        ListNode* head=p.top();
        ListNode* res=head;
        p.pop();
        while(!p.empty()){
            if(head->next)
                p.push(head->next);
            head->next=p.top();
            p.pop();
            head=head->next;
        }
        return res;
    }
};