此題難度為 EASY
先說羅馬數字表還滿複雜的，一定要有個對應表
會比較好理解。
這題我解題是用權重值，而權重值也等於羅馬數字轉換成阿拉伯數字的值。
如果 這個字的權重 < 下一個字的權重，代表此數字將為負值
例如：[代表權重值]
IV= [1]<[5] 故1為負值所以算式為 (-1) + 5 = 4
原始題目為下
Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.
Symbol Value I 1 V 5 X 10 L 50 C 100 D 500 M 1000
For example, two is written as II in Roman numeral, just two one’s added together. Twelve is written as, XII, which is simply X + II. The number twenty seven is written as XXVII, which is XX + V + II.
Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:
Ican be placed beforeV(5) andX(10) to make 4 and 9.Xcan be placed beforeL(50) andC(100) to make 40 and 90.Ccan be placed beforeD(500) andM(1000) to make 400 and 900.
Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999.
package leetCode;
import java.util.*;
public class RomanToInteger {
public static void main(String[] args) {
Solution13 sol = new Solution13();
System.out.println(sol.romanToInt("V"));
}
}
class Solution13 {
public int romanToInt(String s) {
int sum=0;
HashMap<Character,Integer> romanNumTable = new HashMap<Character,Integer>();
List<Integer> intList = new ArrayList<Integer>(); //用來存權重
//HashMap 羅馬數字對印表建立
romanNumTable.put('I',1);
romanNumTable.put('V',5);
romanNumTable.put('X',10);
romanNumTable.put('L',50);
romanNumTable.put('C',100);
romanNumTable.put('D',500);
romanNumTable.put('M',1000);
char[] arrayString = s.toCharArray();
for(char i:arrayString) {
for(HashMap.Entry<Character,Integer> m:romanNumTable.entrySet()) {
if(i==m.getKey())
intList.add(m.getValue());//用對應的值，存成權重intList 裡面的值也等於對應數值
}
}
for(int i=0;i<=intList.size()-1;i++) {
if(i==intList.size()-1) //只有一個字 會到最後一個字的情況下
sum += intList.get(i);
else if(intList.get(i)>=intList.get(i+1)) //這個羅馬署自權重已定要比後大 才家
sum += intList.get(i);
else //權重若變小 則是+負的 例如 IV 等於 -1 +5 = 4
sum -= intList.get(i);
}
return sum;
}
}
觀看更多文章請點MRcoding筆記
