给你一个链表数组，每个链表都已经按升序排列。
请你将所有链表合并到一个升序链表中，返回合并后的链表。
输入：lists = [[1,4,5],[1,3,4],[2,6]]
输出：[1,1,2,3,4,4,5,6]
解释：链表数组如下：
[
1->4->5,
1->3->4,
2->6
]
将它们合并到一个有序链表中得到。
1->1->2->3->4->4->5->6
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* mergeKLists(vector<ListNode*>& lists) {
return merge(lists, 0, lists.size()-1);
}
ListNode* merge(vector<ListNode*>& a, int left, int right) {
if (left == right) return a[left];
if (left > right) return nullptr;
int mid = (right+left)>>1;
return mergeTwoLists(merge(a, left, mid), merge(a, mid+1, right));
}
ListNode* mergeTwoLists(ListNode* a, ListNode* b) {
if (a == nullptr) return b;
if (b == nullptr) return a;
ListNode* head = new ListNode;
ListNode* dummy = head;
while (a && b) {
if (a->val < b->val) {
dummy->next = a;
a = a->next;
} else {
dummy->next = b;
b = b->next;
}
dummy = dummy->next;
}
dummy->next = a != nullptr ? a : b;
return head->next;
}
};
