LeetCode: MergekSortedLists

Title: 

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

排好序的，然后merge，很容易让人联想到归并排序。可以将k个序列按照二分进行分割，然后当长度为1时返回一个排好序的序列，最后将两个序列merge

class Solution{
public:
    ListNode* merge(ListNode* list1, ListNode* list2){
        ListNode head(0),*tail = &head;
        while (list1 != NULL && list2 != NULL){
            if (list1->val < list2->val){
                tail->next = list1;
                tail = list1;
                list1 = list1->next;
            }else{
                tail->next = list2;
                tail = list2;
                list2 = list2->next;
            }
        }
        if (list1 != NULL){
            tail->next = list1;
        }
        if (list2 != NULL){
            tail->next = list2;
        }
        return head.next;
    }
    ListNode* divide(int l,int r,vector<ListNode* > &lists){
        int m = (l + r) / 2;
        if (l < r){
            return merge(divide(l,m,lists),divide(m+1,r,lists));
        }else
            return lists[l];
    }
    ListNode* mergeKLists(vector<ListNode* > &lists){
        int k = lists.size();
        if (0 == k)
            return NULL;
        return divide(0,k-1,lists);
    }
};