Leetcode ☞ 12. Integer to Roman ☆

12. Integer to Roman

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Question

Total Accepted: 57002 Total
Submissions: 151392 Difficulty: Medium

Given an integer, convert it to a roman numeral.
Input is guaranteed to be within the range from 1 to 3999.

跟第13题Leetcode ☞ 13. Roman to Integer相反：http://blog.csdn.net/dr_unknown/article/details/50762420

我的AC（22ms，击败77%）：

<span style="font-size:18px;">char* intToRoman(int num) {
char* ans = (char*)calloc(20, sizeof(char));
int index = 0;

while(num >= 1000){
ans[index++] = 'M';//M:1000
num -= 1000;
}
if (num >= 900){
ans[index++] = 'C';//C:100
ans[index++] = 'M';//M:1000
num -= 900;
}
if (num >= 500){
ans[index++] = 'D';//D:500
num -= 500;
}
if(num >= 400){
ans[index++] = 'C';//C:100
ans[index++] = 'D';//D:500
num -= 400;
}

while(num >= 100){
ans[index++] = 'C';//C:100
num -= 100;
}
if (num >= 90){
ans[index++] = 'X';//X:10
ans[index++] = 'C';//C:100
num -= 90;
}
if (num >= 50){
ans[index++] = 'L';//L:50
num -= 50;
}
if(num >= 40){
ans[index++] = 'X';//X:10
ans[index++] = 'L';//L:50
num -= 40;
}

while(num >= 10){
ans[index++] = 'X';//X:10
num -= 10;
}
if (num >= 9){
ans[index++] = 'I';//I:1
ans[index++] = 'X';//X:10
num -= 9;
}
if (num >= 5){
ans[index++] = 'V';//V:5
num -= 5;
}
if(num >= 4){
ans[index++] = 'I';//I:1
ans[index++] = 'V';//V:5
num -= 4;
}

while(num){
ans[index++] = 'I';
num--;
}

ans[index] = '\0';
return ans;
}</span>
分析：

1、4是5之前减一个1，8是5后面加三个1，9是10之前减一个1。知道这些就能知道罗马数字是如何组合的了。

2、此解法是很细的分情况。还有另外两种解法，时间一样：

C语言：https://leetcode.com/discuss/89381/three-simple-understanding-different-solutions-explained