Merge k Sorted Lists

Problem:

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

采用分治来做。两两合并链表，直到只剩下1个链表。时间复杂度为O(m*n*log(n))，而顺序合并的时间复杂度为O(m*n^2)。

Solution:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *mergeKLists(vector<ListNode *> &lists) {
        if(lists.size()==0)
         return NULL;
       int size = lists.size();
       while(size>1)
       {
         for(int i=0;i<size/2;i++)
              mergeTwo(lists.at(i),lists.at(size-i-1));
         size = (size+1)/2;
        }
        return lists.at(0);
    }
private:
    void mergeTwo(ListNode *&n1,ListNode *&n2)
    {
         ListNode *p = (ListNode*)malloc(sizeof(ListNode)), *q = (ListNode*)malloc(sizeof(ListNode));
         p->next = n1;
         n1 = p;
         q->next = n2;
         n2 = q;
         while(NULL!=p->next&&NULL!=q->next)
         {
              while(NULL!=p->next&&p->next->val<q->next->val)p = p->next;
              ListNode *tmp = p->next;
              p->next = q->next;
              q->next = q->next->next;
              p->next->next = tmp;
              p = p->next;
         }
         if(NULL==p->next)
              p->next = q->next;
         free(n2);
         n2 = n1;
         n1 = n1->next;
         free(n2);
    }
};
​