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Integer to Roman
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(); }
Integer to Roman
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