[LeetCode]12.Integer to Roman

标签：leetcode   数组   经典面试题   

【题目】

Given an integer, convert it to a roman numeral.

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

【分析】

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

还有一些特殊的：每两个阶段的之间有一个减法的表示，比如900=CM， C写在M前面表示M-C。

求商得到每个罗马文字的个数（如：3999  / 1000 = 3 结果有3个M） 

【代码】

/*********************************
*   日期：2015-01-21
*   作者：SJF0115
*   题目: 12.Integer to Roman
*   网址：https://oj.leetcode.com/problems/integer-to-roman/
*   结果：AC
*   来源：LeetCode
*   博客：
**********************************/
#include <iostream>
using namespace std;

class Solution {
public:
    string intToRoman(int num) {
        string result;
        string roman[] = {"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"};
        int value[] = {1000,900,500,400,100,90,50,40,10,9,5,4,1};
        int count;
        // 转换为罗马数字
        for(int i = 0;i < 13;++i){
            count = num / value[i];
            result += toRoman(count,roman[i]);
            num = num % value[i];
        }//if
        return result;
    }
private:
    string toRoman(int num,string str){
        string result;
        for(int i = 0;i < num;++i){
            result += str;
        }//for
        return result;
    }
};

int main(){
    Solution solution;
    int num = 3999;
    string result = solution.intToRoman(num);
    // 输出
    cout<<result<<endl;
    return 0;
}

[LeetCode]12.Integer to Roman

标签：leetcode   数组   经典面试题   

原文地址：http://blog.csdn.net/sunnyyoona/article/details/42971467