LeetCode 23. 合并K个排序链表 【时间击败98.35%】 【内存击败94.11%】

 

原本是“逐一两两合并链表”，用时118ms，换成假分治后只用4ms，震惊！代码其实就是把遍历换成了队列……

 

关键代码变化：

 

“分治”：

static public ListNode mergeKLists(ListNode[] lists) {
        if (lists == null || lists.length == 0) return null;
        Queue<ListNode> q = new ArrayDeque<>();
        for (ListNode n : lists) if (n != null) q.add(n);
        while (q.size() > 1) {
            q.add(merge2(q.poll(), q.poll()));
        }
        return q.poll();
    }

 

逐一两两合并链表：

    public ListNode mergeKLists(ListNode[] lists) {
        if (lists.length==0)return null;
        ListNode head = lists[0];
        for (int i = 1; i < lists.length; i++) head = merge2(head, lists[i]);
        return head;
    }

有毒……

import java.util.ArrayDeque;
import java.util.Queue;

public class Solution {
    static public ListNode mergeKLists(ListNode[] lists) {
        if (lists == null || lists.length == 0) return null;
        Queue<ListNode> q = new ArrayDeque<>();
        for (ListNode n : lists) if (n != null) q.add(n);
        while (q.size() > 1) {
            q.add(merge2(q.poll(), q.poll()));
        }
        return q.poll();
    }

    static ListNode merge2(ListNode l1, ListNode l2) {
        if (l1 == null) return l2;
        if (l2 == null) return l1;

        if (l1.val > l2.val) {
            ListNode t = l1;
            l1 = l2;
            l2 = t;
        }

        ListNode head = l1;
        while (l1 != null && l2 != null) {
            while (l1.next != null && l1.next.val < l2.val) l1 = l1.next;
            if (l1.next == null) {
                l1.next = l2;
                return head;
            }
            ListNode next = l1.next;
            l1.next = l2;
            while (l2.next != null && l2.next.val < next.val) l2 = l2.next;
            ListNode t = l2;
            l2 = l2.next;
            t.next = next;
            l1 = next;
        }
        return head;
    }
}