23. Merge k Sorted Lists

You are given an array of k linked-lists lists, each linked-list is sorted in ascending order.

Merge all the linked-lists into one sorted linked-list and return it.

Example 1:

Input: lists = [[1,4,5],[1,3,4],[2,6]]
Output: [1,1,2,3,4,4,5,6]
Explanation:23. Merge k Sorted Lists 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

Example 2:

Input: lists = []
Output: []

Example 3:

Input: lists = [[]]
Output: []

Solution:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        ListNode dummy = new ListNode(-1);
        ListNode cur = dummy;
        if (lists == null || lists.length == 0){
            return dummy.next;
        }
        PriorityQueue<ListNode> pq = new PriorityQueue<ListNode>((a, b) -> (a.val - b.val));

        for (int i = 0; i < lists.length; i++){
            if (lists[i] != null){            
                pq.offer(lists[i]);
            }

        }

        while(!pq.isEmpty()){
            ListNode pqCur = pq.poll();

            cur.next = pqCur;
            cur = pqCur;
            if (pqCur.next != null){
                pq.offer(pqCur.next);
            }
        }

        return dummy.next;


    }
}

// TC: O(nlogn)
// SC: O(n)

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        ListNode dummy = new ListNode(-1);

        PriorityQueue<ListNode> pq = new PriorityQueue<ListNode>((a,b) -> Integer.compare(a.val, b.val));

        for (int i = 0; i < lists.length; i++){
            ListNode cur = lists[i];
            if (cur != null){
                pq.offer(cur);
            }
        }

        ListNode curr = dummy;

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

        return dummy.next;
    }
}
// TC: O(n)
// SC: O(n)

直接条件反射了

看到 K ... -> priorityQueue

思路对, 但经常@Override 格式啥的 记不住

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        if (lists == null || lists.length == 0){
            return null;
        }

        PriorityQueue<ListNode> minHeap = new PriorityQueue<ListNode>(new Comparator<ListNode>(){
            @Override
            public int compare(ListNode one, ListNode two){
                if (one.val < two.val){
                    return -1;
                }else if (one.val == two.val){
                    return 0;
                }else{
                    return 1;
                }
            }
        });

        for (int i = 0; i < lists.length; i++){
            if (lists[i] != null){
                minHeap.offer(lists[i]);
            }

        }

        ListNode dummy = new ListNode(-1);
        ListNode cur = dummy;
        while(!minHeap.isEmpty()){
            cur.next = minHeap.poll();
            if (cur.next.next != null){
                minHeap.offer(cur.next.next);
            }
            cur = cur.next;
        }

        return dummy.next;
    }
}

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        PriorityQueue<ListNode> minHeap = new PriorityQueue<ListNode>(11, new MyComparator());


        for(ListNode node : lists){
            if (node != null){
                minHeap.offer(node);
            }
        }


        ListNode dummy = new ListNode(0);
        ListNode cur = dummy;

        while(!minHeap.isEmpty()){
            cur.next = minHeap.poll();
            if (cur.next.next != null){
                minHeap.offer(cur.next.next);
            }

            cur = cur.next;
        }



        return dummy.next;
    }

    static class MyComparator implements Comparator<ListNode> {
        public int compare(ListNode o1, ListNode o2){
            if (o1.val == o2.val){
                return 0;
            }else{
                if (o1.val < o2.val){
                    return -1;
                }else{
                    return 1;
                }
            }
        }
    }
}

// TC: O(nlogk + n) = O(nlogk)
// SC: O(k)