题解 | #合并k个已排序的链表#

优先队列，将全部非空的节点全部入队，然后每次选出最小的后，将选出的链表头指针往后移动一个位置，再次入队，如此反复

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *mergeKLists(vector<ListNode *> &lists) {
        auto cmp = [](ListNode* x, ListNode* y){
            return x->val > y->val;
        };
        priority_queue<ListNode*, vector<ListNode*>, decltype(cmp)> q(cmp);
        for(ListNode* l : lists){
            if(l) q.push(l);
        }
        ListNode* dummy = new ListNode(0);
        ListNode* cur = dummy;
        while(!q.empty()){
            ListNode* tmp = q.top();
            q.pop();
            cur->next = tmp;
            cur = cur->next;
            tmp = tmp->next;
            if(tmp) q.push(tmp);
        }
        return dummy->next;
    }
};