23. Merge k Sorted Lists

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

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
/*归并两子链表*/
    ListNode* mergeTwoLists(ListNode* L1, ListNode* L2){
        if(L1 == NULL)
            return L2;
        if(L2 == NULL)
            return L1;
        if(L1->val <= L2->val){
            L1->next = mergeTwoLists(L1->next, L2);
            return L1;
        }
        else{
            L2->next = mergeTwoLists(L1, L2->next);
            return L2;
        }
    }
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        if(lists.empty())
            return NULL;
        ListNode* subList;
        while(lists.size()>1){
            subList = mergeTwoLists(lists[0], lists[1]);//每次将lists的前两个子链表归并
            lists.push_back(subList);//归并后的链表插到lists末尾
            /*删除lists中前两个子链表，直到lists最后只剩一个链表*/
            lists.erase(lists.begin());
            lists.erase(lists.begin());
        }//while
        return lists.front();//返回最后的链表
    }
};