Java实现 LeetCode 23 合并K个排序链表

23. 合并K个排序链表

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

著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

PS：直接用PriorityQueue自动排序，改写一下compare方法。

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

        if (lists.length == 0) {
            return null;
        }

        ListNode dummyHead = new ListNode(0);
        ListNode curr = dummyHead;
        PriorityQueue<ListNode> pq = new PriorityQueue<>(new Comparator<ListNode>() {
            @Override
            public int compare(ListNode o1, ListNode o2) {
                return o1.val - o2.val;
            }
        });

        for (ListNode list : lists) {
            if (list == null) {
                continue;
            }
            pq.add(list);
        }

        while (!pq.isEmpty()) {
            ListNode nextNode = pq.poll();
            curr.next = nextNode;
            curr = curr.next;
            if (nextNode.next != null) {
                pq.add(nextNode.next);
            }
        }
        return dummyHead.next;
    }
}

PS：分治

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
  public ListNode mergeKLists(ListNode[] lists){
        if(lists.length == 0)
            return null;
        if(lists.length == 1)
            return lists[0];
        if(lists.length == 2){
           return mergeTwoLists(lists[0],lists[1]);
        }

        int mid = lists.length/2;
        ListNode[] l1 = new ListNode[mid];
        for(int i = 0; i < mid; i++){
            l1[i] = lists[i];
        }

        ListNode[] l2 = new ListNode[lists.length-mid];
        for(int i = mid,j=0; i < lists.length; i++,j++){
            l2[j] = lists[i];
        }

        return mergeTwoLists(mergeKLists(l1),mergeKLists(l2));

    }
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null) return l2;
        if (l2 == null) return l1;

        ListNode head = null;
        if (l1.val <= l2.val){
            head = l1;
            head.next = mergeTwoLists(l1.next, l2);
        } else {
            head = l2;
            head.next = mergeTwoLists(l1, l2.next);
        }
        return head;
    }
}