Merge k Sorted Lists

https://leetcode.com/problems/merge-k-sorted-lists/description/

Merge k sorted linked lists and return it as one sorted list.

Analyze and describe its complexity.

Thoughts

每次遍历找范围内最小，用min heap。每次找到最小pop后再把最小的next push进去作为候选。C++的pq 用greater作为comparator才是min heap，和Java的相反。。

Code

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        PriorityQueue<ListNode> pq = new PriorityQueue<>((a, b) -> a.val - b.val);
        for (int i = 0; i < lists.length; i++) {
            if (lists[i] != null) {
                pq.offer(lists[i]);    
            }
        }

        ListNode dummy = new ListNode(0), node = dummy;

        while (!pq.isEmpty()) {
            ListNode newNode = pq.poll();
            node.next = newNode;
            node = newNode;
            if (newNode.next != null) {
                pq.offer(newNode.next);
            }
        } 

        return dummy.next;
    }
}

/*
 * @lc app=leetcode id=23 lang=cpp
 *
 * [23] Merge k Sorted Lists
 */
/**
 * 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) {
        const auto cmp = [](ListNode *a, ListNode *b) {
            return (a->val) > (b->val);
        };
        priority_queue<ListNode*, vector<ListNode*>, decltype(cmp)> pq(cmp);
        for (const auto node : lists) {
            if (node != nullptr) pq.push(node);
        }
        ListNode *dummy = new ListNode(-1), *cur = dummy;
        while (!pq.empty()) {
            const auto t = pq.top(); pq.pop();
            cur->next = t;
            cur = cur->next;
            if (t->next != nullptr) pq.push(t->next);
        }
        cur->next = nullptr;
        return dummy->next;
    }
};

Aanlysis

TC: O(nlgn)