integer to roman leetcode c++实现

最新推荐文章于 2025-09-12 17:31:31 发布

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最新推荐文章于 2025-09-12 17:31:31 发布

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CC 4.0 BY-SA版权

文章标签：数据结构与算法 c/c++

原文链接：http://www.cnblogs.com/chess/p/4682143.html

本文介绍了一种将整数转换为罗马数字的算法实现，通过建立映射表并使用栈来构造罗马数字字符串，确保转换准确且高效。适用于1至3999范围内的整数。

Given an integer, convert it to a roman numeral.

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

将整数用罗马数字表示。

思路分析：　　

{"","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"}

　　　　首先从输入的整数开始，（1）取其个位上的数值，判定后得到一个字符串（这个字符串也应该位于罗马数字的最右边，所以再用一个栈来解决）

　　　　　　　　　　　　　　　（2）然后将整数除以10（轮到十位上的数值了），再去执行（1）

代码如下：

 1 #include<stack>
 2 class Solution {
 3 public:
 4     string intToRoman(int num) {
 5         vector<vector<string>> data={
 6             {"","I","II","III","IV","V","VI","VII","VIII","IX"},
 7             {"","X","XX","XXX","XL","L","LX","LXX","LXXX","XC"},
 8             {"","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"},
 9             {"","M","MM","MMM"}
10         };
11         stack<string> sta;
12         int i=0;
13         string res="",result="";
14         while(num!=0){
15             int remain=num%10;
16             res=data[i][remain];
17             sta.push(res);
18             ++i;
19             num/=10;
20         }
21         while(!sta.empty()){
22             result+=sta.top();
23             sta.pop();
24         }
25         return result;
26     }
27 };