LeetCode Integer To Roman
Problem
Given an integer, convert it to a roman numeral.
Input is guaranteed to be within the range from 1 to 3999.
跟Roman To Integer是对应问题。即将整数转换成对应的罗马数字。罗马数字规则见wiki：罗马数字规则
Java 实现（个人解法）
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package com.coderli.leetcode.algorithms.medium;
/**
* Given an integer, convert it to a roman numeral.
* <p>
* Input is guaranteed to be within the range from 1 to 3999
*
* @author li.hzh
*/
public class IntegerToRoman {
public static void main(String[] args) {
IntegerToRoman integerToRoman = new IntegerToRoman();
System.out.println(integerToRoman.intToRoman(1));
System.out.println(integerToRoman.intToRoman(10));
System.out.println(integerToRoman.intToRoman(90));
System.out.println(integerToRoman.intToRoman(1437));
}
public String intToRoman(int num) {
int times = 1;
String result = "";
while (num > 0) {
int tempIntValue = num % 10;
num /= 10;
String tempStrValue = "";
if (tempIntValue == 5) {
switch (times) {
case 1:
result = tempStrValue = "V";
break;
case 2:
result = "L" + result;
break;
case 3:
result = "D" + result;
break;
}
} else if (tempIntValue % 5 <= 3) {
int length = tempIntValue;
switch (times) {
case 1:
if (tempIntValue > 5) {
tempStrValue = "V";
length = tempIntValue - 5;
}
for (int i = 0; i < length; i++) {
tempStrValue += "I";
}
result = tempStrValue + result;
break;
case 2:
if (tempIntValue > 5) {
tempStrValue = "L";
length = tempIntValue - 5;
}
for (int i = 0; i < length; i++) {
tempStrValue += "X";
}
result = tempStrValue + result;
break;
case 3:
if (tempIntValue > 5) {
tempStrValue = "D";
length = tempIntValue - 5;
}
for (int i = 0; i < length; i++) {
tempStrValue += "C";
}
result = tempStrValue + result;
break;
case 4:
for (int i = 0; i < tempIntValue; i++) {
tempStrValue += "M";
}
result = tempStrValue + result;
break;
}
} else {
switch (times) {
case 1:
result = tempIntValue < 5 ? "IV" + result : "IX" + result;
break;
case 2:
result = tempIntValue < 5 ? "XL" + result : "XC" + result;
break;
case 3:
result = tempIntValue < 5 ? "CD" + result : "CM" + result;
break;
}
}
times++;
}
return result;
}
}
分析
上面解法其实受到了罗马转整数问题的惯性思维。把思维限制到了罗马数字只能用基础的I、V、X等元素上。罗马数字的规则不再介绍了。思路其实跟罗马转数字一致，抓住罗马数字其实也是个位、十位、百位等一位一位组合起来的，因此我们一位一位的生成对应的罗马数字即可。
正因为罗马数字也是一位一位对应的，因此网上流传一个简便的写法，代码很简洁：
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public static String change(int num) {
String [][]str =
{
{"","I","II","III","IV","V","VI","VII","VIII","IX"},
{"","X","XX","XXX","XL","L","LX","LXX","LXXX","XC"},
{"","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"},
{"","M","MM","MMM"}
};
String roman = (str[3][num / 1000 % 10])+(str[2][num / 100 % 10])+(str[1][num / 10 % 10])+(str[0][num % 10]);
return roman;
把个十百千各位可能罗马字符准备好，然后直接计算整数对应位上的值，取出数组中对应索引位的罗马值拼到一起就可以了。
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