本文是在学习中的总结,欢迎转载但请注明出处:http://blog.csdn.net/pistolove/article/details/42744649


Given an integer, convert it to a roman numeral.

Input is guaranteed to be within the range from 1 to 3999.

思路:

(1)题意为给定任意1—3999的整数,将其转化为罗马数字。

(2)该题和将罗马数字转化为整数类似,详见罗马数字转化为整数

(3)由于技术有限,本文还是先运用“暴力破解”的思想,对于1-10、10-100、100-1000、1000-3999分别进行讨论,由于比较简单,这里就不累赘了,详情见下方代码。

(4)希望本文对你有所帮助。

算法代码实现如下:

	/**
	 * @author liqq
	 */
    public String intToRoman(int num) {
    	Map<Integer, String> maps = new HashMap<Integer, String>();
    	maps.put(1,"I");
    	maps.put(2,"II");
    	maps.put(3,"III");
    	maps.put(4,"IV");
    	maps.put(5,"V");
    	maps.put(6,"VI");
    	maps.put(7,"VII");
    	maps.put(8,"VIII");
    	maps.put(9,"IX");
    	maps.put(10,"X");
    	maps.put(20,"XX");
    	maps.put(30,"XXX");
    	maps.put(40,"XL");
    	maps.put(50,"L");
    	maps.put(60,"LX");
    	maps.put(70,"LXX");
    	maps.put(80,"LXXX");
    	maps.put(90,"XC");
    	maps.put(100,"C");
    	maps.put(200,"CC");
    	maps.put(300,"CCC");
    	maps.put(400,"CD");
    	maps.put(500,"D");
    	maps.put(600,"DC");
    	maps.put(700,"DCC");
    	maps.put(800,"DCCC");
    	maps.put(900,"CM");
    	maps.put(1000,"M");
    	maps.put(2000,"MM");
    	maps.put(3000,"MMM");

    	StringBuffer buffer = new StringBuffer();
    	if(num<=10){
    		return maps.get(num);
    	}else if(num>10 && num<100){
    		int index = num/10;
    		buffer.append(maps.get(index*10));
    		if(num-index*10>0){
    			buffer.append(maps.get(num-index*10));
    		}
    		return buffer.toString();
    	}else if(num>=100 && num<1000){
    		int hun = num/100;
    		buffer.append(maps.get(hun*100));
    		int te = (num-hun*100)/10;
    		if(te>0){
    			buffer.append(maps.get(te*10));
    		}
    		if(num-hun*100-te*10>0){
        		buffer.append(maps.get(num-hun*100-te*10));
    		}
    		return buffer.toString();
    	}else if(num>=1000 &&num<=3999){
    		int th = num/1000;
    		buffer.append(maps.get(th*1000));
    		int hun = (num-th*1000)/100;
    		if(hun>0){
    			buffer.append(maps.get(hun*100));
    		}
    		int te = (num-th*1000-hun*100)/10;
    		if(te>0){
    			buffer.append(maps.get(te*10));
    		}
    		if(num-th*1000-hun*100-te*10>0){
        		buffer.append(maps.get(num-th*1000-hun*100-te*10));
    		}
    		return buffer.toString();
    	}
    	return buffer.toString();
    }

Leetcode_12_Integer to Roman的更多相关文章

  1. [LeetCode] Roman to Integer 罗马数字转化成整数

    Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 t ...

  2. [LeetCode] Integer to Roman 整数转化成罗马数字

    Given an integer, convert it to a roman numeral. Input is guaranteed to be within the range from 1 t ...

  3. 【leetcode】Roman to Integer

    题目描述: Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range fr ...

  4. [Leetcode] Roman to Integer

    Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 t ...

  5. Integer to Roman -- LeetCode 012

    Given an integer, convert it to a roman numeral. Input is guaranteed to be within the range from 1 t ...

  6. 【LeetCode】Roman to Integer & Integer to Roman

    Roman to Integer Given a roman numeral, convert it to an integer. Input is guaranteed to be within t ...

  7. No.013:Roman to Integer

    问题: Given a roman numeral, convert it to an integer.Input is guaranteed to be within the range from ...

  8. 【leetcode】Integer to Roman

    Integer to Roman Given an integer, convert it to a roman numeral. Input is guaranteed to be within t ...

  9. 【leetcode】Integer to Roman & Roman to Integer(easy)

    Roman to Integer Given a roman numeral, convert it to an integer. Input is guaranteed to be within t ...

随机推荐

  1. POSIX 消息队列相关问题

    一.查看和删除消息队列要想看到创建的posix消息队列,需要在root用户下执行以下操作:# mkdir /dev/mqueue# mount -t mqueue none /dev/mqueue删除 ...

  2. 无法启动postgresql的错误

    chown postgres /etc/ssl/private/ssl-cert-snakeoil.key chgrp postgres /etc/ssl/private/ssl-cert-snake ...

  3. leetcode 之 Single Number II

    问题来源:Single Number II 问题描述:给定一个整数数组,除了一个整数出现一次之外,其余的每一个整数均出现三次,请找出这个出现一次的整数. 大家可能很熟悉另一个题目(Single Num ...

  4. Apache Curator入门实战

    Apache Curator入门实战 Curator是Netflix公司开源的一个Zookeeper客户端,与Zookeeper提供的原生客户端相比,Curator的抽象层次更高,简化了Zookeep ...

  5. Device Mapper 代码分析

    Device Mapper(DM)是Linux 2.6全面引入的块设备新构架,通过DM可以灵活地管理系统中所有的真实或虚拟的块设备. DM以块设备的形式注册到Linux内核中,凡是挂载(或者说&quo ...

  6. Swift中的可选协议和方法的历史渊源

    @objc protocol Transaction { func commit() -> Bool optional func isComplete() -> Bool } 以上协议被标 ...

  7. Android开发使用Java8新特性

    Android 支持所有 Java 7 语言功能,以及一部分 Java 8 语言功能(具体因平台版本而异).本文介绍您可以使用的新语言功能.如何正确配置项目以使用这些功能,以及您可能遇到的任何已知问题 ...

  8. Mybatis源码分析--返回值ResultType和ResultMap

    这一篇博客我们来介绍一下Mybatis执行sql语句返回的结果值的到实体对象的映射机制.首先ResultType和ResultMap的使用方式是不同的. ResultType的使用方式: result ...

  9. Hibernate通过SQL查询常量时只能返回第一个字符的解决方法

    在Hibernate中如果通过 [java] view plaincopy session.createSQLQuery("select '合计' as name from dual&quo ...

  10. Android Demo---实现从底部弹出窗口

    在前面的博文中,小编简单的介绍了如何制作圆角的按钮以及圆角的图片,伴着键盘和手指之间的舞步,迎来新的问题,不知道小伙伴有没有这样的经历,以App为例,点击头像的时候,会从底部弹出一个窗口,有从相册中选 ...