[leetcode 12] Integer to Roman

题目：

Given an integer, convert it to a roman numeral.

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

思路：

1.题目要求将输入的正数用罗马数字表示，由于题目设定给定的数小于4000，那么该问题就比较容易解决；

2.设立4个string数组，分别存储个位到千位的罗马数字；

3.函数中先对int的千位进行处理，再对百位进行处理（x/100)，用获得的数与相应string数组中的值做比对，依次类推，获得该为的罗马数字，直到处理完个位；

4.思路简单，可直接看代码。

代码：

class Solution{public:string intToRoman(int num){string bit[9]={"I","II","III","IV","V","VI","VII","VIII","IX"};string dec[9]={"X","XX","XXX","XL","L","LX","LXX","LXXX","XC"};string hun[9]={"C","CC","CCC","CD","D","DC","DCC","DCCC","CM"};string kbit="M";int k=num/1000;string result="";for(int i=0;i!=k;++i)result += kbit;num=num%1000;int h=num/100;if(h>0)result+=hun[h-1];num=num%100;int d=num/10;if(d>0)result+=dec[d-1];num=num%10;int b=num/1;if(b>0)result+=bit[b-1];return result;}};