本周的题目是一道经典题,难度级别"Medium"...

题目:将1~3999的阿拉伯数字转化为罗马数字。

这题还是比较经典的,做这题之前要先了解一下罗马数字,然后看看实现代码:

char* intToRoman(int num) {
    int numArr[] = {1000,900,500,400,100,90,50,40,10,9,5,4,1};
    char *numRoman[] = {"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"};
    char *numResult = malloc(100);
    
    for(int i=0; i<sizeof(numArr)/sizeof(int); i++){
        while(num >= numArr[i]){
            strcat(numResult,numRoman[i]);
            num -= numArr[i];
        }
    }
    return numResult;
}

百度百科的罗马数字里面就有这道经典的算法题，不过里面是用C++写的，在这里我也翻译成C，顺便用了个for循环，不易出错:

char* intToRoman(int num) 
    char* c[4][10]={
        {"","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"}
    };
    char *roman = malloc(100);
    for (int i = 3; i >= 0; i--) {
        int temp = pow(10, i);
        strcat(roman, Roman[i][num / temp % 10]);
    }
}

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