leetcode--Merge k Sorted Lists

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

method 1: use PriorityQueue (simple)

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode mergeKLists(List<ListNode> lists) {
        ListNode newHead = new ListNode(0);
		ListNode currentNode = newHead;
		if(lists.size() > 0){
			PriorityQueue<ListNode> nodes = new PriorityQueue<ListNode>(lists.size(), new Comparator<ListNode>(){
			    	public int compare(ListNode l, ListNode n){
		                return Integer.compare(l.val, n.val);    
                    }    
			});
			for(int i = 0; i < lists.size(); ++i)
			    if(lists.get(i) != null) 
				    nodes.add(lists.get(i));
			
			while(!nodes.isEmpty()){
				ListNode nodeToSort = nodes.poll();
				if(nodeToSort.next != null)
					nodes.add(nodeToSort.next);
				currentNode.next = nodeToSort;
				currentNode = currentNode.next;
			}
		}
		return newHead.next;
    }
}

　　

　　

method 2: merge 2 lists each step.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode mergeKLists(ArrayList<ListNode> lists) {
        ListNode head = null;
		if(!lists.isEmpty()){
			while(lists.size() > 1){
				ListNode first = lists.remove(0);
				ListNode second = lists.remove(0);
				ListNode newNode = merge2Lists(first, second);
				lists.add(0, newNode);
			}
		    head = lists.get(0);
		}
		return head;
    }
	
	public ListNode merge2Lists(ListNode first, ListNode second){
		ListNode head = new ListNode(0);
		ListNode nextNode = head;
		while(first != null && second != null){
			if(first.val <= second.val){
				nextNode.next = first;
				first = first.next;
				nextNode = nextNode.next;
			}
			else{
				nextNode.next = second;
				second = second.next;
				nextNode = nextNode.next;
			}
		}
		if(first == null)
			nextNode.next = second;
		else
			nextNode.next = first;
		head = head.next;
		return head;
    }
}