题目：

class Solution {
public:
    //合并两个升序链表
    ListNode* mergetwo(ListNode* l1, ListNode* l2){
        if(!l1 || !l2) return l1 ? l1 : l2;
        ListNode* dummy = new ListNode(0);
        ListNode* cur = dummy;
        while(l1&&l2){
            if(l1->val < l2->val){
                cur->next = l1;
                l1 = l1->next;
            }else{
                cur->next = l2;
                l2 = l2->next;
            }
            cur = cur->next;
        }
        cur->next = l1 ? l1 : l2;
        return dummy->next;
    }
    //分治递归合并
    ListNode* merge(vector<ListNode*>& lists, int l, int r){
        if(l==r) return lists[l];       //只剩一个链表
        if(l>r) return nullptr;     //传入空链表的情况
        int mid = (l+r)/2;
        ListNode* left = merge(lists, l, mid);      //合并前半部分链表
        ListNode* right = merge(lists, mid+1, r);       //合并后半部分链表
        return mergetwo(left, right);
    }

    ListNode* mergeKLists(vector<ListNode*>& lists) {
        return merge(lists, 0, lists.size()-1);
    }
};