roman-to-integer
/**
*
* @author gentleKay
* Given a roman numeral, convert it to an integer.
* Input is guaranteed to be within the range from 1 to 3999.
*
* 给定一个罗马数字,将其转换为整数。
* 输入保证在1到3999之间。
*/
这两题可以对比的进行学习:
integer-to-roman: https://www.cnblogs.com/strive-19970713/p/11238129.html
/** * * @author gentleKay * Given a roman numeral, convert it to an integer. * Input is guaranteed to be within the range from 1 to 3999. * * 给定一个罗马数字,将其转换为整数。 * 输入保证在1到3999之间。 */ public class Main14 { public static void main(String[] args) { String s = "I"; System.out.println(Main14.romanToInt(s)); } public static int romanToInt(String s) { char[] ch = s.toCharArray(); char[] roman = {'M','D','C','L','X','V','I'}; int[] num = {1000, 500, 100, 50, 10, 5, 1}; int sum = 0; for (int i=0;i<ch.length;i++) { for (int j=0;j<roman.length;j++) { if (ch[i] == roman[j]) { sum = sum + num[j]; break; } } } //"IV","IX"}; 4 , 9 || 5-1 , 10-1 //"XL","XC"}; 40, 90 || 50-10, 100-10 //"CD","CM"}; 400,900 || 500-100,1000-100
// 主要注意上面的6种情况,需要后面的数 - 前面的数, 因为上面的循环统一都是加上去了,所以现在这里减的话要 乘2. for (int i=0;i<ch.length-1;i++) { if (ch[i]=='I' && ch[i+1]=='V') { sum = sum - 2; } if (ch[i]=='I' && ch[i+1]=='X') { sum = sum - 2; } if (ch[i]=='X' && ch[i+1]=='L') { sum = sum - 20; } if (ch[i]=='X' && ch[i+1]=='C') { sum = sum - 20; } if (ch[i]=='C' && ch[i+1]=='D') { sum = sum -200; } if (ch[i]=='C' && ch[i+1]=='M') { sum = sum -200; } } return sum; } }