【LeetCode算法練習（C++）】Merge k Sorted Lists

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

鏈接：Merge k Sorted Lists
解法：遞歸實現歸併，利用merge Two Lists，每次合併兩個序列。時間O(nlogn)

class Solution {
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        if (lists.size() == 0) return NULL;
        else return VmergeKLists(lists)[0];
    }

    vector<ListNode*> VmergeKLists(vector<ListNode*>& lists) {
        vector<ListNode*> res;
        if (lists.size() == 1) return lists;
        else if (lists.size() == 2) {
            res.push_back(mergeTwoLists(lists[0], lists[1]));
            return res;
        }
        else {
            vector<ListNode*> vl;
            for (int i = 1; i < lists.size(); i += 2) {
                vl.push_back(mergeTwoLists(lists[i - 1], lists[i]));
            }
            if (lists.size() % 2) vl.push_back(lists[lists.size() - 1]);
            return VmergeKLists(vl);
        }
    }

    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        ListNode *head = new ListNode(0);
        head->next = NULL;
        ListNode *p = head;
        while (l1 || l2) {
            ListNode *tmp = new ListNode(0);
            tmp->next = NULL;
            if (!l1 && l2) {
                tmp->val = l2->val;
                l2 = l2->next;
            } else if (!l2 && l1) {
                tmp->val = l1->val;
                l1 = l1->next;
            } else if (l1->val > l2->val) {
                tmp->val = l2->val;
                l2 = l2->next;
            } else {
                tmp->val = l1->val;
                l1 = l1->next;
            }
            p->next = tmp;
            p = p->next;
        }
        p = head;
        head = head->next;
        p->next = NULL;
        delete(p);
        return head;
    }
};

Runtime: 32 ms