题解 | #合并k个已排序的链表#

合并k个已排序的链表

https://www.nowcoder.com/practice/65cfde9e5b9b4cf2b6bafa5f3ef33fa6

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 (lists == null || lists.size() == 0) {
            return null;
        }
        ListNode newHead = new ListNode(-1);
        newHead.next = lists.get(0);
        ListNode temp = newHead.next;
        for (int i = 1; i < lists.size(); i++) {
            ListNode node = newHead;
            ListNode cur = lists.get(i);
            while (temp != null && cur != null) {
                if (temp.val < cur.val) {
                    node.next = temp;
                    temp = temp.next;
                } else {
                    node.next = cur;
                    cur = cur.next;
                }
                node = node.next;
            }
            if (temp != null) {
                node.next = temp;
            }
            if (cur != null) {
                node.next = cur;
            }
            temp = newHead.next;
        }
        return newHead.next;
    }
}

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