合并K个排序的列表，不适用于边缘情况，如null，并且只有2个列表C++

#include <iostream>
#include <vector>
#include <queue>

using namespace std;

struct ListNode {
    int val;
    ListNode* next;
    ListNode(int x) : val(x), next(NULL) {}
};

struct Compare {
    bool operator()(const ListNode* a, const ListNode* b) {
        return a->val > b->val;
    }
};

ListNode* mergeKLists(vector<ListNode*>& lists) {
    priority_queue<ListNode*, vector<ListNode*>, Compare> pq;
    for (ListNode* list : lists) {
        if (list) {
            pq.push(list);
        }
    }

    ListNode* dummy = new ListNode(0);
    ListNode* tail = dummy;
    while (!pq.empty()) {
        ListNode* curr = pq.top();
        pq.pop();
        tail->next = curr;
        tail = tail->next;
        if (curr->next) {
            pq.push(curr->next);
        }
    }

    return dummy->next;
}

int main() {
    // 示例用法
    vector<ListNode*> lists;
    ListNode* list1 = new ListNode(1);
    list1->next = new ListNode(4);
    list1->next->next = new ListNode(5);

    ListNode* list2 = new ListNode(1);
    list2->next = new ListNode(3);
    list2->next->next = new ListNode(4);

    ListNode* list3 = new ListNode(2);
    list3->next = new ListNode(6);

    lists.push_back(list1);
    lists.push_back(list2);
    lists.push_back(list3);

    ListNode* mergedList = mergeKLists(lists);

    while (mergedList) {
        cout << mergedList->val << " ";
        mergedList = mergedList->next;
    }
    cout << endl;

    return 0;
}