LeetCode 12 整数转罗马数字 java

方法一
用一个大While循环通过各种if-else语句依次减少num的值构造str，虽然时间代价不大但是代码冗长：

class Solution {
public String intToRoman(int num) {
String str = "";
while(num>0){
if(num>=1000){
num-=1000;
str+='M';
}
else if(num >= 900&&num<1000){
num-=900;
str+="CM";
}
else if(num>=500&&num<900){
num-=500;
str+="D";
}
else if(num>=400&&num<500){
num-=400;
str+="CD";
}
else if(num>=100&&num<400){
num-=100;
str+="C";
}
else if(num>=90&&num<100){
num-=90;
str+="XC";
}
else if(num>=50&&num<90){
num-=50;
str+="L";
}
else if(num>=40&&num<50){
num-=40;
str+="XL";
}
else if(num>=10&&num<40){
num-=10;
str+="X";
}
else if(num>=9&&num<10){
num-=9;
str+="IX";
}
else if(num>=5&&num<9){
num-=5;
str+="V";
}
else if(num==4){
num-=4;
str+="IV";
}
else{
num-=1;
str+="I";
}
}
return str;
}
}

可以通过将各种不同大小的数字与对应的罗马字母依次存到两个数组里，用查表的方式遍历数组，可精简代码量：

class Solution {
public String intToRoman(int num) {
int values[]=new int[]{1000,900,500,400,100,90,50,40,10,9,5,4,1};
String strs[]=new String[]{"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"};

String str="";
for(int i=0; i<13; i++){
if(num==0)
break;
int times=num/values[i];
if(times==0)
continue;
for(int t=0;t<times;t++)
{
str+=strs[i];
}
num=num-(times*values[i]);
}
return str;
}
}