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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 };
integer to roman leetcode c++实现
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原文地址：http://www.cnblogs.com/chess/p/4682143.html
