题解 | #合并k个已排序的链表#

import java.util.*;

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 *   public ListNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param lists ListNode类ArrayList 
     * @return ListNode类
     */
    public ListNode mergeKLists (ArrayList<ListNode> lists) {
        // write code here
        int n = lists.size();
        ListNode head = new ListNode(0);
        ListNode cur = head;
        ListNode inf = new ListNode(1001);
	  	//记录每次遍历minNode的索引j
        int j = 0;
	  	//初始化minNode
        ListNode minNode = inf;
        while(true){
            for(int i=0;i<n;i++){
                if (lists.get(i) != null){
                    if (minNode.val > lists.get(i).val){
					  //遍历过程中不断更新minNode与j
                        minNode = lists.get(i);
                        j = i;
                    }
                }
            }
            if(minNode == inf){
			  //最后一次遍历,minNode还是初始值,说明所有节点都已经被添加到head中,终止循环
                break;
            }else{
			  //更新lists[j],将minnode添加到cur后面
                lists.set(j,minNode.next);
                cur.next = minNode;
                cur = cur.next;
			  //初始化minNode
                minNode = inf;
            }
        }
        return head.next;
    }
}

用一个minNode跟一个指针j记录每次遍历lists得到的最小的Node,把得到的minNode放在cur后面,之后把第j个链表的头指针后移,并初始化minNode,这样循环往复,直到所有的链表节点被遍历完,break循环.