合併 k 個排序連結串列，返回合併後的排序連結串列。請分析和描述演算法的複雜度。

示例:

輸入:
[
  1->4->5,
  1->3->4,
  2->6
]
輸出: 1->1->2->3->4->4->5->6

分析：先合併兩個有序連結串列，然後再將先前合併的與剩餘的連結串列繼續兩兩合併。時間複雜度為O(logn)

/**
 * 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) {
        if(lists.empty()) {
            return NULL;
        }
        int n = lists.size();
        while(n > 1) {
            int k = (n + 1) / 2;
            for(int i = 0; i < n / 2; i++) {
                lists[
 
i] = merge2List(lists[i], lists[i + k]);
            }
            n = k;
        }
        return lists[0];
    }
    ListNode* merge2List(ListNode* n1, ListNode* n2) {
        ListNode dummy(0);
        ListNode* p = &dummy;
        while(n1 && n2) {
            if(n1->val < n2->
 
val) {
                p->next = n1;
                n1 = n1->next;
            } else {
                p->next = n2;
                n2 = n2->next;
            }
            p = p->next;
        }
        if(n1) {
            p->next = n1;
        } else if(n2) {
            p->next = n2;
        }
        return dummy.next;
    }
};