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) {}
 * };
 */
#include <queue>
struct Compare {
    bool operator() (ListNode* p, ListNode* q) {
        return p->val > q->val;
    }
};
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
     */
     priority_queue<int> res;
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        // write code here
        priority_queue<ListNode*, vector<ListNode*>, Compare> pq;
        for (int i = 0; i < lists.size(); ++i) {
            if(lists[i])
                pq.push(lists[i]);
        }
        ListNode node(-1),*p = &node;
        while (!pq.empty()) {
            ListNode * t = pq.top();
            pq.pop();
            p->next = t;
            p = t;
            if(t->next!=nullptr)
            {
                pq.push(t->next);
            }
            p->next = nullptr;
        }
        return node.next;
    }
};