方法一，一位一位判断。
class Solution {
public String intToRoman1(int num) {
StringBuilder sb = new StringBuilder();
char ooo[] = new char[]{'I','V','X','L','C','D','M'};
int wei = 0;
while(num>0){
int n = num %10;
if(n<4){
for(int i=0;i<n;i++){
sb.append(ooo[wei*2]);
}
}else if(n==4){
sb.append(ooo[wei*2+1]);
sb.append(ooo[wei*2]);
}else if(n==5){
sb.append(ooo[wei*2+1]);
}else if(n>5 && n<9){
for(int i=0;i<n-5;i++){
sb.append(ooo[wei*2]);
}
sb.append(ooo[wei*2+1]);
}else if(n==9){
sb.append(ooo[wei*2+1+1]);
sb.append(ooo[wei*2]);
}
wei++;
num = num/10;
}
return sb.reverse().toString();
}
方法二，因为知道最多4位数，所以打表暴力。
public String intToRoman(int num) {
String s1[] = new String[]{"","I","II","III","IV","V","VI","VII","VIII","IX"};
String s2[] = new String[]{"","X","XX","XXX","XL","L","LX","LXX","LXXX","XC"};
String s3[] = new String[]{"","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"};
String s4[] = new String[]{"","M","MM","MMM"};
// return s4[(num/1000)%10]+s3[(num/100)%10]+s2[(num/10)%10]+s1[num%10];
return new StringBuilder().append(s4[(num/1000)]).append(s3[(num/100)%10]).append(s2[(num/10)%10]).append(s1[num%10]).toString();
}
}