leetcode Problem12 Integer to Roman

Given an integer, convert it to a roman numeral.

Input is guaranteed to be within the range from 1 to 3999.

思路:首先先把num按照个位十位百位千位拆分，然后对每一位上的数字按照roman to integer的对应表格，转化为对应的roman，然后就所有转化的再拼起来就是所求的结果。代码如下(已通过leetcode)

public class Solution {

public String intToRoman(int num) {

String[] str=new String[4];

String str2="";

if(num<1 || num>3999) return null;

int[] a=new int[]{0,0,0,0};

a[0]=num/1000;

if(a[0]!=0) num=num-a[0]*1000;

a[1]=num/100;

if(a[1]!=0) num=num-a[1]*100;

a[2]=num/10;

if(a[2]!=0) num=num-a[2]*10;

a[3]=num;

switch(a[3]) {

case 0: break;

case 1: str[3]="I";

break;

case 2: str[3]="II";

break;

case 3: str[3]="III";

break;

case 4: str[3]="IV";

break;

case 5: str[3]="V";

break;

case 6: str[3]="VI";

break;

case 7: str[3]="VII";

break;

case 8: str[3]="VIII";

break;

case 9: str[3]="IX";

break;

}

switch(a[2]) {

case 0: break;

case 1: str[2]="X";

break;

case 2: str[2]="XX";

break;

case 3: str[2]="XXX";

break;

case 4: str[2]="XL";

break;

case 5: str[2]="L";

break;

case 6: str[2]="LX";

break;

case 7: str[2]="LXX";

break;

case 8: str[2]="LXXX";

break;

case 9: str[2]="XC";

break;

}

switch(a[1]) {

case 0: break;

case 1: str[1]="C";

break;

case 2: str[1]="CC";

break;

case 3: str[1]="CCC";

break;

case 4: str[1]="CD";

break;

case 5: str[1]="D";

break;

case 6: str[1]="DC";

break;

case 7: str[1]="DCC";

break;

case 8: str[1]="DCCC";

break;

case 9: str[1]="CM";

break;

}

switch(a[0]) {

case 0: break;

case 1: str[0]="M";

break;

case 2: str[0]="MM";

break;

case 3: str[0]="MMM";

break;

default: break;

}

for(int i=0;i<4;i++) {

if(str[i]!=null)

str2+=str[i];

}

return str2;

}

}