Merge k Sorted Lists -- leetcode

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

两种实现方法，第一种采用优先队列，第二种采用分治

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    /**     * use priority_queue    class mycomparison    {    public:        bool operator() (ListNode* lhs, ListNode* rhs)        {                return (lhs->val > rhs->val);        }    };        ListNode *mergeKLists(vector<ListNode*> &lists) {                priority_queue<ListNode*, vector<ListNode *>, mycomparison>  queue;        for(int i = 0; i < lists.size(); i++){            if(lists[i] != NULL)                queue.push(lists[i]);        }                ListNode* head =NULL;        ListNode* prev = NULL;        while(!queue.empty())            {            ListNode* tmp = queue.top();            queue.pop();            if(!prev)            {                head = tmp;                prev = tmp;            }            else            {                prev->next= tmp;                prev = tmp;            }            if(tmp->next)                queue.push(tmp->next);                    }        return head;    }    */    //use divide and conquer algorithm    //merge two list    ListNode* Merge(ListNode* list1, ListNode* list2)    {        ListNode* head = NULL;        if(list1 == NULL)            return list2;        if(list2 == NULL)            return list1;        if(list1->val < list2->val){            head = list1;            head->next = Merge(list1->next, list2);        }            else{            head = list2;            head->next = Merge(list1, list2->next);        }        return head;    }    ListNode* MergeCore(vector<ListNode*> &lists, int left, int right)    {        if(left < right)        {            int mid = (right+left)/2;            return Merge(MergeCore(lists, left, mid), MergeCore(lists, mid+1, right));        }        else if(left == right)            return lists[left];        return NULL;        }    ListNode *mergeKLists(vector<ListNode*> &lists) {        if(lists.size() <= 0)            return NULL;        return MergeCore(lists, 0, lists.size()-1);    }};