LeetCode--Merge k Sorted Lists

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

题解：建立一个大小为k的最小堆，即堆顶存放最小元素，因为k个链表已排好序，借鉴归并排序的技巧，先取出堆顶的元素，再放入堆顶元素所在链表的表头的元素，再维持最小堆循环迭代。
注意直接链表存入会导致逆序，可以先存入vector，再逆序存入链表。

代码如下：

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    int insertNode(ListNode** head,int val)    {        ListNode* newNode = new ListNode(val);        newNode->next = *head;        *head = newNode;        return 0;    }    class qNode{    public:        int val;        int index;        friend bool operator<(qNode a,qNode b){            return a.val > b.val;            //val小的优先级高，先从queue pop出来        }    };    ListNode *mergeKLists(vector<ListNode *> &lists) {        priority_queue<qNode> q;        vector<int> tmp;        qNode now,next;        ListNode* head = NULL;        for(int i = 0;i < lists.size();i++){            if(lists[i] != NULL){                now.val = lists[i]->val;                lists[i] = lists[i]->next;                now.index = i;                q.push(now);            }        }        while(!q.empty()){            now = q.top();            q.pop();            tmp.push_back(now.val);            if(lists[now.index] != NULL){                next.val = lists[now.index]->val;                next.index = now.index;                q.push(next);                lists[now.index] = lists[now.index]->next;            }        }        for(int i = tmp.size()-1;i >= 0;i--)        {            insertNode(&head,tmp[i]);        }        return head;    }};