一. 问题描述
罗马数字包含以下七种字符: I, V, X, L,C,D 和 M。
字符 数值
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
ps:
I 可以放在 V (5) 和 X (10) 的左边,来表示 4 和 9。
X 可以放在 L (50) 和 C (100) 的左边,来表示 40 和 90。
C 可以放在 D (500) 和 M (1000) 的左边,来表示 400 和 900。
给定一个整数,将其转为罗马数字。输入确保在 1 到 3999 的范围内。
示例 1:
输入: 3
输出: "III"
示例 2:
输入: 4
输出: "IV"
示例 3:
输入: 9
输出: "IX"
示例 4:
输入: 58
输出: "LVIII"
解释: L = 50, V = 5, III = 3.
示例 5:
输入: 1994
输出: "MCMXCIV"
解释: M = 1000, CM = 900, XC = 90, IV = 4.
二. 问题分析
对于给定的数据,我们需要从最高位向最低位判断即可,利用递归的方式重复调用就可以得到结果.
三. 代码书写
自己写的(low了点)
public String intToRoman(int num) {
String result = "";
int I = 1;
int V = 5;
int X = 10;
int L = 50;
int C = 100;
int D = 500;
int M = 1000;
if (num >= M) {
if (num < 4 * M) {
for (int i=0; i<num/M; i++) {
result += "M";
}
int surplus = num % M;
result += intToRoman(surplus);
} else {
return "输入超限";
}
} else if (num >= D){
if (num < M - C) {
result += "D";
for (int i=0;i<(num-D)/C; i++) {
result += "C";
}
} else {
result += "CM";
}
int surplus = num % C;
result += intToRoman(surplus);
} else if (num >= C){
if (num < 4 * C) {
for (int i=0; i<num/C; i++) {
result += "C";
}
} else {
result += "CD";
}
int surplus = num % C;
result += intToRoman(surplus);
} else if (num >= L) {
if (num < C - X) {
result += "L";
for (int i=0; i<(num-L)/X; i++) {
result += "X";
}
} else {
result += "XC";
}
int surplus = num % X;
result += intToRoman(surplus);
} else if (num >= X) {
if (num < 4 * X) {
for (int i=0; i<num/X; i++) {
result += "X";
}
} else {
result += "XL";
}
int surplus = num % X;
result += intToRoman(surplus);
} else if (num >= V) {
if (num < X - I) {
result += "V";
for (int i=0; i<(num-V); i++) {
result += "I";
}
}else {
result += "IX";
}
} else if (num >= I) {
if (num < V -I) {
for (int i=0;i<num; i++) {
result += "I";
}
} else {
result += "IV";
}
}
return result;
}
牛人解答
public String intToRomanBM(int num) {
String[][] c = new String[][]{
{"","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 = "";
roman += c[3][num / 1000];
roman += c[2][num / 100 % 10];
roman += c[1][num / 10 % 10];
roman += c[0][num % 10];
return roman;
}
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