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
     */
    ListNode* mergeKLists(vector<ListNode*>& lists){
        auto cmp=[&](ListNode *a,ListNode *b){return a->val>b->val;};
        priority_queue<ListNode*,vector<ListNode*>,decltype(cmp)>heap(cmp);
        for(auto &x:lists)if(x!=nullptr)heap.push(x);
        if(heap.empty())return nullptr;
        ListNode *h=heap.top();
        ListNode *pre=h;
        heap.pop();
        if(pre->next!=nullptr)heap.push(pre->next);
        while(!heap.empty()){
            ListNode *cur=heap.top();
            heap.pop();
            pre->next=cur;
            pre=cur;
            if(cur->next!=nullptr)heap.push(cur->next);
        }
        return h;
    }
};