[12] Integer to Roman

© Jarvus Chen / www.jarvus.net

Description

Given an integer, convert it to a roman numeral.

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

Required knowledge

Roman numerals conversion table

• I = 1
• V = 5
• X = 10
• L = 50
• C = 100
• D = 500
• M = 1000

String array

https://stackoverflow.com/questions/1088622/how-do-i-create-an-array-of-strings-in-c

There are three way you can declare a string array. The second one is the most recommended.

char a1[][14] = { "blah", "hmm" };
char* a2[] = { "blah", "hmm" };
char (*a3[])[] = { &"blah", &"hmm" };

// prints blah
printf(a1[0]); 
printf(a2[0]);
printf(*a3[0]);

// change its content
strcpy(a1[0], "hmm");
strcpy(a1[1], "blah");

a2[0] = "hmm";
a2[1] = "blah";

a3[0] = &"hmm";
a3[1] = &"blah";

String connect

https://openhome.cc/Gossip/CGossip/StringLengthCopyCat.html

You can select how many char you want to combine with another one.

char str1[80] = "xyz";
char str2[] = "abc";

strcat(str1, str2);
// str1 is xyzabc
strncat(str1, str2, 2);
// str1 is xyzabcab

My solution

Part 1 identify how many digit in the input number and reverse it. Because of strcatmethod, I have to process roman transformation from the last digit.

Part 2 according to remain values, choose which data can it retrieve.

char* intToRoman(int num) {

    char* data1[] = {"I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"};
    char* data2[] = {"X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"};
    char* data3[] = {"C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"};
    char* data4[] = {"M", "MM", "MMM"};

    char* ans = malloc(sizeof(char)*15);
    // Part 1
    // inverse int
    int temp = 0;
    int count = 0;
    while( num > 0){
        temp = temp * 10 + num % 10;
        num /= 10;
        count++;
    };
    // Part 2
    // process from the last digit
    while ( temp > 0){

        int digit = temp%10;
        if ( count == 4 && digit > 0){
            strcat(ans, data4[digit-1]);
        }else if( count == 3 && digit > 0){
            strcat(ans, data3[digit-1]);
        }else if( count == 2 && digit > 0){
            strcat(ans, data2[digit-1]);
        }else if( count == 1 && digit > 0){
            strcat(ans, data1[digit-1]);
        }
        temp = temp/10;
        count--;
    }
    return ans;

}

Others solution

https://discuss.leetcode.com/topic/11001/my-c-solution-in-19ms

Not really good but this is kind of another thinking logic to finish the problem.

char *link(char *s1,char *s2)
{
    int len1=strlen(s1);
    int len2=strlen(s2);
    for(int i=len1;i<len1+len2;i++)
    s1[i] = s2[i-len1];
    return s1;
}

char *integerToRoman(int x)
{
    char *arr[13]={"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"};
    int value[13]= {1000,900,500,400,100,90,50,40,10,9,5,4,1};
    char roman=(char)malloc(sizeof(char)*100);
    memset(roman,0,100);
    for(int i=0;x!=0;i++){
        while(x>=value[i]){
            x-=value[i];
            roman = link(roman,arr[i]);
        }    
    }
    return roman;
}