[leetcode] Integer to Roman

最新推荐文章于 2020-06-29 15:21:54 发布

Cabinathor

最新推荐文章于 2020-06-29 15:21:54 发布

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分类专栏： Leetcode

本文链接：https://blog.youkuaiyun.com/lmxmimihuhu/article/details/46429089

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本文介绍了一种将整数转换为罗马数字的算法实现，输入范围限制在1到3999之间。通过两个不同版本的代码示例展示了如何使用C++来实现这一转换过程。

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From : https://leetcode.com/problems/integer-to-roman/

Given an integer, convert it to a roman numeral.

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

class Solution {
public:
    string intToRoman(int num) {
        if(num<1 || num >3999) return "";
        string ans;
        int cnt, i=0;// i odd:1**  even:5**
        map<int, char> base;
        base[1]='I';
        base[5] = 'V';
        base[10] = 'X';
        base[50] = 'L';
        base[100] = 'C';
        base[500] = 'D';
        base[1000] = 'M';
        for(map<int, char>::reverse_iterator it=base.rbegin(); it!=base.rend(); it++) {
            i++;
            if(num/it->first) {
                cnt = num/it->first;
                num -= cnt*it->first;
                if(i%2==1&&cnt==4) {
                    ans += it->second;
                    it--;
                    ans += it->second;
                    it++;
                    cnt = 0;
                } else if(i%2==0&&cnt==1&&(num+it->first/5)/it->first==1) {
                    num -= 4*it->first/5;
                    it++;
                    ans += it->second;
                    it--;it--;
                    ans += it->second;
                    it++;
                    cnt = 0;
                }
                char ch = it->second;
                while(cnt--) {
                    ans += ch;
                }
            }
        }
        return ans;
    }
};

class Solution {
public:
    string intToRoman(int num) {
        if(num<1 || num >3999) return "";
        string ans;
        int bases[] = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};
        string values[] = {"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"};
        for(int i=0; i<13; i++) {
            while(num >= bases[i]) {
                num -= bases[i];
                ans += values[i];
            }
        }
        return ans;
    }
};