题解 | 合并k个已排序的链表 | prority_queue

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 *	ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
#include <cfloat>
#include <memory>
#include <queue>

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 Comapre
{
bool operator()(struct ListNode* l1,struct ListNode* l2)
    {
        return l1->val  >  l2->val;
    }
};

    ListNode* mergeKLists(vector<ListNode*>& lists) {
       priority_queue<struct ListNode*,vector<struct ListNode*>,Comapre> Q;
       for(auto& l : lists)
       {
        if(l != nullptr)
        Q.push(l);
       }
       auto* dummy = new struct ListNode(-1);
       auto* ptr = dummy;
       while(Q.size())
       {
        ptr ->next = Q.top(); ptr = ptr ->next;
        if(Q.top() ->next != nullptr)
        Q.push(Q.top() ->next);
        Q.pop();
       }
    ptr = dummy ->next;
    delete dummy;
    return ptr;
    }
};