题解 | #合并k个已排序的链表#

import java.util.*;
/**
 * 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) {
        if (null == lists || lists.size() == 0) return null;
        ListNode head = null;
        // 向前走的指针
        ListNode minNode = null;
        int minIndex = -1;
        for (int i = 0; i < lists.size(); i++) {
            ListNode node = lists.get(i);
            if (null != node) {
                if (null == minNode) {
                    minNode = node;
                    minIndex = i;
                } else {
                    // 判断当前遍历的节点的值是否小于minNode
                    if (minNode.val > node.val) {
                        minNode = node;
                        minIndex = i;
                    }
                }
            }
        }
        if (minNode == null) return null;
        lists.set(minIndex, minNode.next);
        head = minNode;
        head.next = mergeKLists(lists);
        return head;
    }
}