Merge k Sorted Lists

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */bool operator <(const pair<ListNode*,int> &p1, const pair<ListNode*,int> &p2){    return p1.first->val < p2.first->val;}/*假设投针数组为副本，且由其它地方销毁*/#include<queue>class Solution {public:    ListNode* mergeKLists(vector<ListNode*>& lists) {        ListNode* helpNode = new ListNode(0);        ListNode* cur = helpNode;        typedef pair<ListNode*,int> heaptype;        priority_queue<heaptype,vector<heaptype>,greater<heaptype>> q;//最小队列        for(int i = 0; i< lists.size(); ++i){            if(lists[i]){                q.push(make_pair(lists[i],i));                lists[i] = lists[i]->next;            }        }        while(!q.empty()){            ListNode* node = q.top().first;            int index = q.top().second;            q.pop();            if(lists[index]){                q.push(make_pair(lists[index],index));                lists[index] = lists[index]->next;            }            cur->next = new ListNode(node->val);            cur= cur->next;        }                auto ret = helpNode->next;        delete helpNode;        return ret;    }};