import java.util.ArrayList;
import java.util.List;
import java.util.PriorityQueue; /**
* Source : https://oj.leetcode.com/problems/merge-k-sorted-lists/
*
* Created by lverpeng on 2017/7/11.
*
* Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
*
*/
public class MergeKList { /**
* 两两归并排序
*
* @param headList
* @return
*/
public Node merge (List<Node> headList) {
while (headList.size() > 1) {
Node list1 = headList.remove(headList.size() - 1);
Node list2 = headList.remove(headList.size() - 1);
Node mergeList = mergeTwoList(list1, list2);
headList.add(0, mergeList);
}
return headList.get(0); } /**
* 对k个链表做归并排序,比较k个元素的时候使用最小堆排序
*
* @param headList
* @return
*/
public Node merge0 (List<Node> headList) {
PriorityQueue<Node> priorityQueue = new PriorityQueue<Node>();
for (Node node : headList) {
priorityQueue.add(node);
} Node head = null;
Node current = null;
while (priorityQueue.size() > 0) {
Node node = priorityQueue.poll();
if (head == null) {
head = node;
current = node;
} else {
if (current == node) {
throw new IllegalArgumentException("出现了循环引用,可能是多个链表中有相同的元素");
}
current.next = node;
current = current.next;
}
if (node.next != null) {
priorityQueue.add(node.next);
}
}
return head;
} /**
* 归并排序
*
* @param list1
* @param list2
* @return
*/
public static Node mergeTwoList (Node list1, Node list2) {
Node head = null;
Node current = null;
while (list1 != null && list2 != null) {
Node node = null;
if (list1.value > list2.value) {
node = list2;
list2 = list2.next;
} else {
node = list1;
list1 = list1.next;
}
if (head == null) {
head = current = node;
} else {
current.next = node;
current = current.next;
}
} list1 = list1 == null ? list2 : list1;
if (list1 != null) {
if (head != null) {
current.next = list1;
} else {
head = list1;
}
}
return head;
} private static class Node implements Comparable<Node>{
int value;
Node next; @Override
public String toString() {
return "Node{" +
"value=" + value +
", next=" + (next == null ? "" : next.value) +
'}';
} @Override
public int compareTo(Node o) {
return this.value - o.value;
}
} private static void print (Node node) {
while (node != null) {
System.out.println(node);
node = node.next;
}
} public static void main(String[] args) {
Node list1 = new Node();
Node pointer1 = list1;
list1.value = 1;
Node list2 = new Node();
list2.value = 2;
Node pointer2 = list2;
Node list3 = new Node();
list3.value = 3;
Node pointer3 = list3;
for (int i = 4; i < 20; i++) {
Node node = new Node();
node.value = i;
if (i % 3 == 1) {
pointer1.next = node;
pointer1 = pointer1.next;
} else if (i % 3 == 2) {
pointer2.next = node;
pointer2 = pointer2.next;
} else {
pointer3.next = node;
pointer3 = pointer3.next;
}
} List<Node> list = new ArrayList<Node>();
list.add(list1);
list.add(list1);
list.add(list1);
MergeKList mergeKList = new MergeKList();
// Node result = mergeKList.merge(list);
// print(result); System.out.println(); Node result = mergeKList.merge0(list);
print(result); }
}

leetcode — merge-k-sorted-lists的更多相关文章

  1. LeetCode: Merge k Sorted Lists 解题报告

    Merge k Sorted Lists Merge k sorted linked lists and return it as one sorted list. Analyze and descr ...

  2. [LeetCode] Merge k Sorted Lists 合并k个有序链表

    Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 这 ...

  3. LeetCode:Merge k Sorted Lists

    题目链接 Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexi ...

  4. LeetCode——Merge k Sorted Lists

    Discription: Merge k sorted linked lists and return it as one sorted list. Analyze and describe its ...

  5. leetcode -- Merge k Sorted Lists add code

    Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. [ ...

  6. LeetCode Merge k Sorted Lists (链表)

    题意 Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity ...

  7. [Leetcode] Merge k sorted lists 合并k个已排序的链表

    Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 思 ...

  8. Leetcode:Merge k Sorted Lists分析和实现

    题目大意是传入一个链表数组lists,每个链表都由若干个链接的链表结点组成,并且每个链表结点记录一个整数.题目保证传入的链表中的整数按从小到大进行排序. 题目要求我们输出一个新的链表,这个链表中应该包 ...

  9. LeetCode Merge k Sorted Lists 解决报告

    https://oj.leetcode.com/problems/merge-k-sorted-lists/ 归并K已经整理阵列,和分析算法的复杂. 解决报告:无论是不考虑优化,最简单的实现是要重新走 ...

  10. LeetCode —— Merge k Sorted Lists

    /* ** 算法的思路: ** 1.将k个链表的首元素进行建堆 ** 2.从堆中取出最小的元素,放到链表中 ** 3.如果取出元素的有后续的元素,则放入堆中,若没有则转步骤2,直到堆为空 */ #in ...

随机推荐

  1. TUN/TAP编程实现

    其实关于这两种设备的编程,基本上属于八股文,大家一般都这么干. 启动设备之前 有的linux 并没有将tun 模块编译到内核之中,所以,我们要做的第一件事情就是检查我们的系统是否支持 TUN/TAP ...

  2. centos7 编译安装nginx+tcp转发

    一.依赖 1. gcc 安装安装 nginx 需要先将官网下载的源码进行编译,编译依赖 gcc 环境,如果没有 gcc 环境,则需要安装: yum install gcc-c++ 2. PCRE pc ...

  3. python运算符优先级

    下面这个表给出Python的运算符优先级,从最低的优先级(最松散地结合)到最高的优先级(最紧密地结合).这意味着在一个表达式中,Python会首先计算表中较下面的运算符,然后在计算列在表上部的运算符. ...

  4. Linux下设置快捷键

    以设置终端为例,进入Settings>>Keyboard>>Custom Shortcuts,点左下脚的+号,Name栏填入Treminal,command栏填入gnome-t ...

  5. JSON笔记整理

    JSON简介:     JSON: JavaScript Object Notation(JavaScript 对象表示法) JSON 是存储和交换文本信息的语法.类似 XML. JSON 比 XML ...

  6. OpenCV基础知识介绍

    1.图像与矩阵 一般来说,图像是一个标准的矩形,有着宽度(width)和高度(height).而矩阵有着行(row)和列(column),矩阵的操作在数学和计算机中的处理都很常见且成熟,于是很自然的就 ...

  7. 《pyhton语言程序设计》_第一章笔记

    章1.62 (1).python区分大小写. (2).python忽略在符号#之后的同行的内容 (3).python和matlab很相似(个人感觉) (4).章节1.91: >>>i ...

  8. 仿boost::any的泛型指针类any的实现

    在boost库中,any是一种特殊容器,只能容纳一个元素,但这个元素可以是任意的类型----int.double.string.标准容器或者任何自定义类型.程序可以用any保存任意的数据,也可以在任何 ...

  9. ubuntu下chrome浏览器安装flash插件(pepperflashplugin-nonfree)

    安装前说明: ubuntu的Google 已经不能使用Adobe Flash了,需要用PepperFlashPlayer来替代 Adobe Flash才行. 安装步骤: 1.安装pepperflash ...

  10. iOS中的HTTPS

    在WWDC 15,Apple提出的ATS (App Transport Security) ,是 Apple 在推进网络通讯安全的一个重要方式.在 iOS 9 和 OS X 10.11 中,默认情况下 ...