integer to roman leetcode c++实现-白红宇

integer to roman leetcode c++实现

阅读量：6246 次

发布时间：2019-06-22

本文共 1314 字，大约阅读时间需要 4 分钟。

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
     
       2 class Solution { 3 public: 4     string intToRoman(int num) { 5         vector
      
       
        > 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
        
          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 };