Merge k Sorted Lists

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

A: 像mergesort一样，两两merge。

递归版本：

    ListNode *merge(ListNode *list1,ListNode *list2)
    {
        if(!list1)
            return list2;
        if(!list2)
            return list1;
        ListNode *head;
        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 *mergeKLists(vector<ListNode *> &lists) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        ListNode *head = NULL;
        for(int i=0;i<lists.size();i++)
        {
            head = merge(head,lists[i]);
        }
        
        return head;
        
        
    }

　　

 迭代版本：

  ListNode *merge(ListNode *list1,ListNode *list2)
    {
        if(!list1)
            return list2;
        if(!list2)
            return list1;
        ListNode *head = NULL,*it1,*it2,*prev,*cur;
        it1 = list1;
        it2 = list2;
        while(it1&&it2)
        {
            if(it1->val<=it2->val)
            {
                cur = it1;
                it1 = it1->next;
            }else{
                cur = it2;
                it2 = it2->next;
            }
            if(!head)
            {
                head = prev = cur; 
            }else
            {
                prev->next = cur;
                prev = cur;
            }
        }
        
        if(it1)
            prev->next = it1;
        
        if(it2)
            prev->next = it2;
            
        return head;
    }
    ListNode *mergeKLists(vector<ListNode *> &lists) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        ListNode *head = NULL;
        for(int i=0;i<lists.size();i++)
        {
            head = merge(head,lists[i]);
        }
        
        return head;
        
        
    }