给定一个罗马数字，任务是找到对应的十进制值。
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例子：
Input: IXOutput: 9IX is a Roman symbol which represents 9 Input: XLOutput: 40XL is a Roman symbol which represents 40Input: MCMIVOutput: 1904M is a thousand, CM is nine hundred and IV is four
罗马数字基于以下符号。
SYMBOL VALUE I 1 IV 4 V 5 IX 9 X 10 XL 40 L 50 XC 90 C 100 CD 400 D 500 CM 900 M 1000
方法： 罗马数字中的数字是按降序书写的符号串（例如，M的首字母，D的后字母，等等）。但是，在一些特定情况下，为了避免四个字符连续重复（如IIII或XXXX）， 减法记数法 常用的用法如下：
- 我 放在 五、 或 十、 表示少了一个，所以是四个 四、 （一个小于5）和9是九（一个小于10）。
- 十、 放在 L 或 C 表示少10个，所以40个是 特大号 （10小于50）且90为 XC （十比一百少）。
- C 放在 D 或 M 表示少了一百，所以四百是 光盘 （一百比五百少）九百是 厘米 （比一千少一百）。
将罗马数字转换为整数的算法：
- 将罗马数字字符串拆分为罗马符号（字符）。
- 将罗马数字的每个符号转换为它所代表的值。
- 从索引0开始逐个获取符号：
- 如果符号的当前值大于或等于下一个符号的值，则将该值添加到运行总数中。
- 否则，将下一个符号的值与运行总数相加，以减去该值。
以下是上述算法的实现：
C++
// Program to convert Roman // Numerals to Numbers #include <bits/stdc++.h> using namespace std; // This function returns value // of a Roman symbol int value( char r) { if (r == 'I' ) return 1; if (r == 'V' ) return 5; if (r == 'X' ) return 10; if (r == 'L' ) return 50; if (r == 'C' ) return 100; if (r == 'D' ) return 500; if (r == 'M' ) return 1000; return -1; } // Returns decimal value of // roman numaral int romanToDecimal(string& str) { // Initialize result int res = 0; // Traverse given input for ( int i = 0; i < str.length(); i++) { // Getting value of symbol s[i] int s1 = value(str[i]); if (i + 1 < str.length()) { // Getting value of symbol s[i+1] int s2 = value(str[i + 1]); // Comparing both values if (s1 >= s2) { // Value of current symbol // is greater or equal to // the next symbol res = res + s1; } else { // Value of current symbol is // less than the next symbol res = res + s2 - s1; i++; } } else { res = res + s1; } } return res; } // Driver Code int main() { // Considering inputs given are valid string str = "MCMIV" ; cout << "Integer form of Roman Numeral is " << romanToDecimal(str) << endl; return 0; } |
JAVA
// Program to convert Roman // Numerals to Numbers import java.util.*; public class RomanToNumber { // This function returns // value of a Roman symbol int value( char r) { if (r == 'I' ) return 1 ; if (r == 'V' ) return 5 ; if (r == 'X' ) return 10 ; if (r == 'L' ) return 50 ; if (r == 'C' ) return 100 ; if (r == 'D' ) return 500 ; if (r == 'M' ) return 1000 ; return - 1 ; } // Finds decimal value of a // given roman numeral int romanToDecimal(String str) { // Initialize result int res = 0 ; for ( int i = 0 ; i < str.length(); i++) { // Getting value of symbol s[i] int s1 = value(str.charAt(i)); // Getting value of symbol s[i+1] if (i + 1 < str.length()) { int s2 = value(str.charAt(i + 1 )); // Comparing both values if (s1 >= s2) { // Value of current symbol // is greater or equalto // the next symbol res = res + s1; } else { // Value of current symbol is // less than the next symbol res = res + s2 - s1; i++; } } else { res = res + s1; } } return res; } // Driver Code public static void main(String args[]) { RomanToNumber ob = new RomanToNumber(); // Considering inputs given are valid String str = "MCMIV" ; System.out.println( "Integer form of Roman Numeral" + " is " + ob.romanToDecimal(str)); } } |
python
# Python program to convert Roman Numerals # to Numbers # This function returns value of each Roman symbol def value(r): if (r = = 'I' ): return 1 if (r = = 'V' ): return 5 if (r = = 'X' ): return 10 if (r = = 'L' ): return 50 if (r = = 'C' ): return 100 if (r = = 'D' ): return 500 if (r = = 'M' ): return 1000 return - 1 def romanToDecimal( str ): res = 0 i = 0 while (i < len ( str )): # Getting value of symbol s[i] s1 = value( str [i]) if (i + 1 < len ( str )): # Getting value of symbol s[i + 1] s2 = value( str [i + 1 ]) # Comparing both values if (s1 > = s2): # Value of current symbol is greater # or equal to the next symbol res = res + s1 i = i + 1 else : # Value of current symbol is greater # or equal to the next symbol res = res + s2 - s1 i = i + 2 else : res = res + s1 i = i + 1 return res # Driver code print ( "Integer form of Roman Numeral is" ), print (romanToDecimal( "MCMIV" )) |
C#
// C# Program to convert Roman // Numerals to Numbers using System; class GFG { // This function returns value // of a Roman symbol public virtual int value( char r) { if (r == 'I' ) return 1; if (r == 'V' ) return 5; if (r == 'X' ) return 10; if (r == 'L' ) return 50; if (r == 'C' ) return 100; if (r == 'D' ) return 500; if (r == 'M' ) return 1000; return -1; } // Finds decimal value of a // given roman numeral public virtual int romanToDecimal( string str) { // Initialize result int res = 0; for ( int i = 0; i < str.Length; i++) { // Getting value of symbol s[i] int s1 = value(str[i]); // Getting value of symbol s[i+1] if (i + 1 < str.Length) { int s2 = value(str[i + 1]); // Comparing both values if (s1 >= s2) { // Value of current symbol is greater // or equalto the next symbol res = res + s1; } else { res = res + s2 - s1; i++; // Value of current symbol is // less than the next symbol } } else { res = res + s1; i++; } } return res; } // Driver Code public static void Main( string [] args) { GFG ob = new GFG(); // Considering inputs given are valid string str = "MCMIV" ; Console.WriteLine( "Integer form of Roman Numeral" + " is " + ob.romanToDecimal(str)); } } // This code is contributed by Shrikant13 |
PHP
<?php // Program to convert Roman // Numerals to Numbers // This function returns // value of a Roman symbol function value( $r ) { if ( $r == 'I' ) return 1; if ( $r == 'V' ) return 5; if ( $r == 'X' ) return 10; if ( $r == 'L' ) return 50; if ( $r == 'C' ) return 100; if ( $r == 'D' ) return 500; if ( $r == 'M' ) return 1000; return -1; } // Returns decimal value // of roman numeral function romanToDecimal(& $str ) { // Initialize result $res = 0; // Traverse given input for ( $i = 0; $i < strlen ( $str ); $i ++) { // Getting value of // symbol s[i] $s1 = value( $str [ $i ]); if ( $i +1 < strlen ( $str )) { // Getting value of // symbol s[i+1] $s2 = value( $str [ $i + 1]); // Comparing both values if ( $s1 >= $s2 ) { // Value of current symbol // is greater or equal to // the next symbol $res = $res + $s1 ; } else { $res = $res + $s2 - $s1 ; $i ++; // Value of current symbol is // less than the next symbol } } else { $res = $res + $s1 ; $i ++; } } return $res ; } // Driver Code // Considering inputs // given are valid $str = "MCMIV" ; echo "Integer form of Roman Numeral is " , romanToDecimal( $str ), "" ; // This code is contributed by ajit ?> |
Python3
def romanToInt(rom): value = { 'M' : 1000 , 'D' : 500 , 'C' : 100 , 'L' : 50 , 'X' : 10 , 'V' : 5 , 'I' : 1 } # Initialize previous character and answer p = 0 ans = 0 # Traverse through all characters n = len (rom) for i in range (n - 1 , - 1 , - 1 ): # If greater than or equal to previous, # add to answer if value[rom[i]] > = p: ans + = value[rom[i]] # If smaller than previous else : ans - = value[rom[i]] # Update previous p = value[rom[i]] print (ans) romanToInt( 'MCMIV' ) |
Javascript
<script> // Program to convert Roman // Numerals to Numberspublic // This function returns // value of a Roman symbol function value(r) { if (r == 'I' ) return 1; if (r == 'V' ) return 5; if (r == 'X' ) return 10; if (r == 'L' ) return 50; if (r == 'C' ) return 100; if (r == 'D' ) return 500; if (r == 'M' ) return 1000; return -1; } // Finds decimal value of a // given roman numeral function romanToDecimal( str) { // Initialize result var res = 0; for (i = 0; i < str.length; i++) { // Getting value of symbol s[i] var s1 = value(str.charAt(i)); // Getting value of symbol s[i+1] if (i + 1 < str.length) { var s2 = value(str.charAt(i + 1)); // Comparing both values if (s1 >= s2) { // Value of current symbol // is greater or equalto // the next symbol res = res + s1; } else { // Value of current symbol is // less than the next symbol res = res + s2 - s1; i++; } } else { res = res + s1; } } return res; } // Driver Code // Considering inputs given are valid var str = "MCMIV" ; document.write( "Integer form of Roman Numeral" + " is " + romanToDecimal(str)); // This code contributed by umadevi9616 </script> |
输出
Integer form of Roman Numeral is 1904
复杂性分析：
- 时间复杂性： O（n），其中n是字符串的长度。 只需要遍历字符串一次。
- 空间复杂性： O（1）。 因为不需要额外的空间。
另一个解决方案——
C++
// Program to convert Roman // Numerals to Numbers #include <bits/stdc++.h> using namespace std; // This function returns value // of a Roman symbol int romanToDecimal(string& str) { map< char , int > m; m.insert({ 'I' , 1 }); m.insert({ 'V' , 5 }); m.insert({ 'X' , 10 }); m.insert({ 'L' , 50 }); m.insert({ 'C' , 100 }); m.insert({ 'D' , 500 }); m.insert({ 'M' , 1000 }); int sum = 0; for ( int i = 0; i < str.length(); i++) { /*If present value is less than next value, subtract present from next value and add the resultant to the sum variable.*/ if (m[str[i]] < m[str[i + 1]]) { sum+=m[str[i+1]]-m[str[i]]; i++; continue ; } sum += m[str[i]]; } return sum; } // Driver Code int main() { // Considering inputs given are valid string str = "MCMIV" ; cout << "Integer form of Roman Numeral is " << romanToDecimal(str) << endl; return 0; } |
JAVA
// Program to convert Roman // Numerals to Numbers import java.util.Map; import java.util.HashMap; class GFG{ private static final Map<Character, Integer> roman = new HashMap<Character, Integer>() {{ put( 'I' , 1 ); put( 'V' , 5 ); put( 'X' , 10 ); put( 'L' , 50 ); put( 'C' , 100 ); put( 'D' , 500 ); put( 'M' , 1000 ); }}; // This function returns value // of a Roman symbol private static int romanToInt(String s) { int sum = 0 ; int n = s.length(); for ( int i = 0 ; i < n; i++) { // If present value is less than next value, // subtract present from next value and add the // resultant to the sum variable. if (i != n - 1 && roman.get(s.charAt(i)) < roman.get(s.charAt(i + 1 ))) { sum += roman.get(s.charAt(i + 1 )) - roman.get(s.charAt(i)); i++; } else { sum += roman.get(s.charAt(i)); } } return sum; } // Driver Code public static void main(String[] args) { // Considering inputs given are valid String input = "MCMIV" ; System.out.print( "Integer form of Roman Numeral is " + romanToInt(input)); } } // This code is contributed by rahuldevgarg |
C#
// Program to convert Roman // Numerals to Numbers using System; using System.Collections.Generic; public class GFG { static Dictionary< char , int > roman = new Dictionary< char , int >(); // This function returns value // of a Roman symbol public static int romanToInt(String s) { int sum = 0; int n = s.Length; for ( int i = 0; i < n; i++) { // If present value is less than next value, // subtract present from next value and add the // resultant to the sum variable. if (i != n - 1 && roman[s[i]] < roman[s[i + 1]]) { sum += roman[s[i + 1]] - roman[s[i]]; i++; } else { sum += roman[s[i]]; } } return sum; } // Driver Code public static void Main(String[] args) { roman[ 'I' ] = 1; roman[ 'V' ] =5; roman[ 'X' ] =10; roman[ 'L' ] =50; roman[ 'C' ] =100; roman[ 'D' ] =500; roman[ 'M' ] =1000; // Considering inputs given are valid String input = "MCMIV" ; Console.Write( "int form of Roman Numeral is " + romanToInt(input)); } } // This code is contributed by Rajput-Ji |
Javascript
<script> // Program to convert Roman // Numerals to Numbers var roman = new Map() ; roman.set( 'I' , 1); roman.set( 'V' , 5); roman.set( 'X' , 10); roman.set( 'L' , 50); roman.set( 'C' , 100); roman.set( 'D' , 500); roman.set( 'M' , 1000); // This function returns value // of a Roman symbol function romanToInt( s) { var sum = 0; var n = s.length; for (i = 0; i < n; i++) { // If present value is less than next value, // subtract present from next value and add the // resultant to the sum variable. if (i != n - 1 && roman.get(s.charAt(i)) < roman.get(s.charAt(i + 1))) { sum += roman.get(s.charAt(i + 1)) - roman.get(s.charAt(i)); i++; } else { sum += roman.get(s.charAt(i)); } } return sum; } // Driver Code // Considering inputs given are valid var input = "MCMIV" ; document.write( "Integer form of Roman Numeral is " + romanToInt(input)); // This code is contributed by Rajput-Ji </script> |
输出
Integer form of Roman Numeral is 1904
时间复杂性 –O（N） 辅助空间 –O（1）
另一个解决方案： 使用python编写更短的代码
Python3
def romanToInt(s): translations = { "I" : 1 , "V" : 5 , "X" : 10 , "L" : 50 , "C" : 100 , "D" : 500 , "M" : 1000 } number = 0 s = s.replace( "IV" , "IIII" ).replace( "IX" , "VIIII" ) s = s.replace( "XL" , "XXXX" ).replace( "XC" , "LXXXX" ) s = s.replace( "CD" , "CCCC" ).replace( "CM" , "DCCCC" ) for char in s: number + = translations[char] print (number) romanToInt( 'MCMIV' ) |
输出
1904
时间复杂性 –O（N）
辅助空间 –O（1）
参考资料： https://en.wikipedia.org/wiki/Roman_numerals 本文由 拉胡尔·阿格拉瓦尔 .如果你喜欢GeekSforgek，并想贡献自己的力量，你也可以使用 写极客。组织 或者把你的文章寄去评论-team@geeksforgeeks.org.看到你的文章出现在Geeksforgeks主页上，并帮助其他极客。 如果您发现任何不正确的地方，或者您想分享有关上述主题的更多信息，请写下评论。
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