LeetCode (力扣) 12. Integer to Roman ( C ) - Medium - 暴力解

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题目为给定一个阿拉伯数字，将之转换成罗马数字的表示方式。

下一题则反过来，把罗马数字转换成阿拉伯数字，此为下一题的连结 : LeetCode 13. Roman to Integer

 

题目与范例如下

 

Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

Symbol Value
I 1
V 5
X 10
L 50
C 100
D 500
M 1000

For example, 2 is written as II in Roman numeral, just two one's added together. 12 is written as XII, which is simply X + II. The number 27 is written as XXVII, which is XX + V + II.

Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:

I can be placed before V (5) and X (10) to make 4 and 9.
X can be placed before L (50) and C (100) to make 40 and 90.
C can be placed before D (500) and M (1000) to make 400 and 900.
Given an integer, convert it to a roman numeral.

Example 1:

Input: num = 3
Output: "III"
Example 2:

Input: num = 4
Output: "IV"
Example 3:

Input: num = 9
Output: "IX"
Example 4:

Input: num = 58
Output: "LVIII"
Explanation: L = 50, V = 5, III = 3.
Example 5:

Input: num = 1994
Output: "MCMXCIV"
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

Constraints:

1 <= num <= 3999

 

题目限制数字范围在 1~3999 之间，故我们的解题策略为透过一系列的 if else 暴力解。

1、先配置回传的记忆体，并宣告一个变量 i 用来当回传值的下标，用以一个个储存罗马数字。

2、透过 while 回圈一个个拆出罗马数字，我们先判断有几个 1000 ，接着因为罗马数字表示 9 与 4 时都是透过减法，把 1 放置于前，代表减一。故每次都要额外检查 9 或 4 开头的数字，以此类推当 num 归零时回传字符串。

下方为我的代码

char * intToRoman(int num){
    char *re = (char*) malloc(sizeof(char)*20);
    int i = 0;
    while(num>0){
        if(num>=1000){
            num-=1000;
            re[i++] = 'M';
        }
        else if(num>=900){
            num-=900;
            re[i++] = 'C';
            re[i++] = 'M';
        }
        else if(num>=500){
            num-=500;
            re[i++] = 'D';
        }
        else if(num>=400){
            num-=400;
            re[i++] = 'C';
            re[i++] = 'D';
        }
        else if(num>=100){
            num-=100;
            re[i++] = 'C';
        }
        else if(num>=90){
            num-=90;
            re[i++] = 'X';
            re[i++] = 'C';
        }
        else if(num>=50){
            num-=50;
            re[i++] = 'L';
        }
        else if(num>=40){
            num-=40;
            re[i++] = 'X';
            re[i++] = 'L';
        }
        else if(num>=10){
            num-=10;
            re[i++] = 'X';
        }
        else if(num>=9){
            num-=9;
            re[i++] = 'I';
            re[i++] = 'X';
        }
        else if(num>=5){
            num-=5;
            re[i++] = 'V';
        }
        else if(num>=4){
            num-=4;
            re[i++] = 'I';
            re[i++] = 'V';
        }
        else{
            num--;
            re[i++] = 'I';
        }
        
        
        
    }
    re[i] = '\0';
    return re;
}

 

下方为时间与空间之消耗

Runtime: 4 ms, faster than 80.79 % of C online submissions for Integer to Roman.

Memory Usage: 5.9 MB, less than 80.46 % of C online submissions for Integer to Roman.