leetCode---Merge k Sorted Lists

一. 题目：Merge k Sorted Lists

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

二. 思路分析

借鉴归并排序的方法，自顶向下，先递归的对链表的前半部分和后半部分进行归并排序，最后再merge。
以下代码顺利AC了，时间复杂度为：O(NlogK),因为递归深度是logK，并且在每个级别中，每个元素将被比较。

/** * 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) {        return partion(lists,0,lists.size()-1);    }public:    ListNode* partion(vector<ListNode*>&lists,int start,int end){                if(start == end) return lists[start];        if(start > end) return NULL;        int mid = (start+end)/2;        ListNode* l1 = partion(lists,start,mid);        ListNode* l2 = partion(lists,mid+1,end);                ListNode* head = new ListNode(0); ListNode* cur = head;        while(l1 !=NULL && l2 != NULL){            if(l1->val < l2->val){                cur->next = l1;                l1 = l1->next;            }else{                cur->next = l2;                l2 = l2->next;            }            cur = cur->next;        }        cur->next = (l1 != NULL)?l1:l2;                return head->next;            }};