23. 合并K个排序链表-链表-困难

问题描述

合并 k 个排序链表，返回合并后的排序链表。请分析和描述算法的复杂度。

示例:

输入:
[
  1->4->5,
  1->3->4,
  2->6
]
输出: 1->1->2->3->4->4->5->6

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/merge-k-sorted-lists

解答

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */

 //遍历list放入数组，再sort生成listnode。10ms
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        int len = lists.length;
        if(len == 0)return null;
        List<Integer> resList = new ArrayList<Integer>();
        for(ListNode temp:lists)
            while(temp!=null){
                resList.add(temp.val);
                temp = temp.next;
            }
        Collections.sort(resList);
        ListNode res = new ListNode(-1), a = res;
        for(int i:resList){
            res.next = new ListNode(i);
            res = res.next;
        }
        //System.out.println(resList);
        return a.next;
    }
}
  /*
 //以下是链表做法，280ms
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        int len = lists.length;
        if(len == 0)return null;
        ListNode res = new ListNode(-1), a = res, min = new ListNode(10000000);
        int position, i;
        while(true){
            position = -1;
            for(i=0;i<len;i++){
                if(lists[i] != null && min.val >= lists[i].val){
                    min = lists[i];
                    position = i;//记录位置
                }
            }
            if(position == -1)return a.next;
            lists[position] = min.next;
            res.next = min;
            res = res.next;
            min = new ListNode(10000000);
        }
    }
}
*/