Reputation: 3
Ques: Given an array of linked-lists lists, each linked list is sorted in ascending order. Merge all the linked-lists into one sort linked-list and return it.
Input: lists = [[1,4,5],[1,3,4],[2,6]]
Output: [1,1,2,3,4,4,5,6]
Explanation: The linked-lists are:
[
1->4->5,
1->3->4,
2->6
]
merging them into one sorted list:
1->1->2->3->4->4->5->6
I don't understand how this custom comparator is working for min-heap priority_queue.
/*
* 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) {
priority_queue<ListNode*, vector<ListNode*>, comp> pq;
for(auto head : lists){
if(head != NULL)
pq.push(head);
}
ListNode* dummy = new ListNode(-1);
ListNode* curr = dummy;
while(!pq.empty()){
curr->next = pq.top();
pq.pop();
curr = curr->next;
ListNode* next = curr->next;
if(next != NULL)
pq.push(next);
}
return dummy->next;
}
struct comp{
bool operator()(ListNode* a, ListNode* b){
return a->val > b->val;
}
};
};
Why the return val is a->val > b->val
instead of a->val < b->val
Upvotes: 0
Views: 292
Reputation: 81926
std::priority_queue is documented on cppreference with:
A user-provided Compare can be supplied to change the ordering, e.g. using
std::greater<T>
would cause the smallest element to appear as thetop()
.
So if you want to have a priority queue where you can pop the smallest element, it expects you to pass a comparator that returns true if a > b
.
(And note that you're leaking the object allocated at dummy
)
Upvotes: 1