leetcode刷题12，13（Integer to Roman&Roman to Integer）

一个条件：int型变量在1到3999之间，不超过罗马数字的表示范围

int转罗马数字的思路：一个二维String数组存储对应的1十进制各个位的0到9表示的罗马数字（比如个位的4是IV，十位的5是L，百位的6是DC）。对输入的int数字按位解析（由低位到高位），然后用二维数组的罗马数字表示,返回的结果是一个字符串不断叠加的过程。

public class Solution {
public String intToRoman(int num) {
String[][] roman = {
{"","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"}
};
String str = "";
int i = 0;
while(num!=0)
{
int yu = num%10;
str = roman[i][yu]+ str;
num = num/10;
i++;
}
return str;
}
}
罗马数字转int的思路：罗马数字这个字符串需要按位解析每个字符（index从0到str.length，对应的十进制从高位到低位）。如果某一位的字符对应的数字大于等于下一位字符对应的数字，那么计算结果加上这一位（本位）字符代表的数字。否则，计算结果减去这一位（本位）字符代表的数字（如IX，IV这样，直接减去1）。最后再把字符串最后一个字符对应的数字加上计算结果。

public class Solution {
public int romanToInt(String s) {
HashMap<Character,Integer> map = new HashMap<Character,Integer>();
map.put('I',1);
map.put('V',5);
map.put('X',10);
map.put('L',50);
map.put('C',100);
map.put('D',500);
map.put('M',1000);
int ans = 0;
char[] str = s.toCharArray();
ans+=map.get(str[str.length-1]);
for(int i=0;i<str.length-1;i++)
{
int val = map.get(str[i]);
int nextval = map.get(str[i+1]);
if(val<nextval)
ans-= val;
else

4000
ans+= val;
}
return ans;

}
}