[LeetCode]023-Merge K Sorted Lists

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

Solution(1):
初始做法，虽使用递归，但其实还是两两合并，算法设计的比较差。超时。
复杂度为O(n^2);

ListNode* mergeKLists(vector<ListNode*>& lists) {        int n =lists.size();        if(n == 0)            return NULL;        merge_all_Lists(lists,n);        return lists[0];    }    void merge_all_Lists(vector<ListNode*>& lists,int k)     {        ListNode* r_list = new ListNode(NULL);        ListNode* head = r_list;        if(k == 0 || k == 1)            return ;        else if(k==2)        {            ListNode* l1 = lists[0];            ListNode* l2 = lists[1];            if(l1 == NULL)            {                lists.erase(lists.begin());                return ;            }            if(l2 == NULL)            {                lists.erase(lists.begin()+1);                return ;            }            while(l1 != NULL && l2 != NULL)            {                if(l1->val < l2->val)                {                    r_list->val = l1->val;                    l1 = l1->next;                }                else                {                    r_list->val = l2->val;                    l2 = l2->next;                }                if(l1 != NULL && l2 != NULL)                {                    r_list->next = new ListNode(NULL);                    r_list = r_list->next;                }            }            while(l1 != NULL)            {                r_list->next = new ListNode(NULL);                r_list = r_list->next;                r_list->val = l1->val;                l1 = l1->next;            }            while(l2 != NULL)            {                r_list->next = new ListNode(NULL);                r_list = r_list->next;                r_list->val = l2->val;                l2 = l2->next;            }            lists.erase(lists.begin());            lists.erase(lists.begin());            lists.insert(lists.begin(),head);            return;        }        else        {            while(lists.size() >1)                merge_all_Lists(lists,k-1);         }    }

Solution(2):
修改下，使用分治法来做，效率就提升了，复杂度为O(nlogn)。

    ListNode* mergeKLists(vector<ListNode*>& lists)     {        int n = lists.size();        if(n == 0)            return NULL;        return merge_lists(lists,0,n-1);    }    ListNode* merge_lists(vector<ListNode*> &lists,int low,int high)    {        if(low<high)        {            int mid = (low+high)/2;            return mergeTwoLists(merge_lists(lists,low,mid),merge_lists(lists,mid+1,high));        }        return lists[low];    }    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2)     {        if(l1 == NULL)            return l2;        if(l2 == NULL)            return l1;        ListNode * t1 = l1;        ListNode * t2 = l2;        ListNode * l3 = new ListNode(NULL);        ListNode *head = l3;        while(t1!=NULL && t2!=NULL)        {            if(t1->val < t2->val)            {                l3->val = t1->val;                t1 = t1->next;            }            else            {                l3->val = t2->val;                t2 = t2->next;            }            if(t1 != NULL && t2 !=NULL)            {                l3->next = new ListNode(NULL);                l3 = l3->next;            }        }        while(t1!=NULL)        {            l3->next = new ListNode(NULL);            l3 = l3->next;            l3->val = t1->val;            t1 = t1->next;        }        while(t2!=NULL)        {            l3->next = new ListNode(NULL);            l3 = l3->next;            l3->val = t2->val;            t2 = t2->next;        }        return head;    }