23. Merge k Sorted Lists

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

=======

合并k个链表形成一个已排序链表

 

思路:

如何合并两个有序链表?经典merge算法:

ListNode *mergeList(ListNode *head1,ListNode *head2){
        ListNode dummy(-1);
        ListNode *h = &dummy;
        while(head1 && head2){
            if(head1->val <= head2->val){
                h->next = head1;
                head1 = head1->next;
            }else{
                h->next = head2;
                head2 = head2->next;
            }
            h = h->next;
        }
        if(head1) h->next = head1;
        if(head2) h->next = head2;
        return dummy.next;
    }

对vector中的链表依次merge

ListNode* mergeKLists(vector<ListNode*>& lists) {
        ///
        if(lists.empty()) return nullptr;

        ListNode *p = lists[0];
        for(int i = 1;i<(int)lists.size();i++){
            p = mergeList(p,lists[i]);
        }
        return p;
    }

=======

这个算法有点问题:会有大量重复的计算,比如vector[0]中的链表就会被merge ( vector.size()-1)次

可以按照k路merge的思路,做一边.