merge-k-sorted-lists[合并链表]

题目描述

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

• 思路
  按顺序合并所有的链表并返回头结点。
  法一：优先队列存储所有的链表头结点，每次取最小值连接并使它的下一个加入到队列中。
  法二：从第一个链表开始依次合并后再返回头结点。

//法一
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
import java.util.*;
public class Solution {
    public ListNode mergeKLists(ArrayList lists) {
        if(lists == null || lists.size() == 0)
            return null;

        PriorityQueue queue = new PriorityQueue(lists.size(), 
                                                                    new Comparator(){
            @Override
            public int compare(ListNode l1, ListNode l2){
                if(l1.val > l2.val)
                    return 1;
                else if(l1.val < l2.val)
                    return -1;
                else
                    return 0;
            }
        });

        ListNode head = new ListNode(0);
        ListNode Head = head;

        for(ListNode list : lists){
            if(list != null)
                queue.add(list);
        }

        while( !queue.isEmpty()){
            head.next = queue.poll();
            head = head.next;
            if(head.next != null)
                queue.add(head.next);
        }

        return Head.next;
    }
}

//法二
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
import java.util.*;
public class Solution {
    public ListNode mergeKLists(ArrayList lists) {
        if(lists == null || lists.size() == 0)
            return null;

        ListNode head = lists.get(0);
        
        for(int i=1; i