Merge k Sorted Lists

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

思路：这道题是归并排序的加强版。设置一个数组Min，记录每个链表的待归并的最小值，再设置一个变量minum为所有链表待归并的值的最小值。假设待归并第i个链表时，只需将链表的值与minum对比，若小于，继续向后归并。每归并n个链表后，再找出最小的minum。时间复杂度为O(MN)，其中M为链表数，N为最长的链表长度。空间复杂度为O(M)。

class Solution {public:    ListNode *mergeKLists(vector<ListNode *> &lists) {        int len = lists.size();        if(len == 0)        {            return NULL;        }        int i,j;        vector<int> Min;        Min.resize(len);        int empty = 0;        int minum = INT_MAX;        ListNode* head = new ListNode(0);        ListNode* p = head;        for(i=0; i<len; ++i)        {            if(lists[i] == NULL)            {                Min[i] = INT_MAX;            }                else             {                Min[i] = lists[i]->val;            }              if(minum > Min[i])            {                minum = Min[i];            }              }         while(empty < len)  //链表全为空         {            empty = 0;            for(i=0; i<len; ++i)            {                while(lists[i] != NULL && lists[i]->val <= minum)                {                    ListNode* pNode = new ListNode(lists[i]->val);                    p->next = pNode;                    p = p->next;                    lists[i] = lists[i]->next;                }                    if (lists[i] != NULL)                {                    Min[i] = lists[i]->val;                }                    else                {                    empty++;                    Min[i] = INT_MAX;                }                }             minum = INT_MAX;            for(i=0; i<len; ++i)            {                if(minum > Min[i])                {                    minum = Min[i];                }                }               }         return head->next;          }};