给定一个 长度为N 的字符串S ，如果给定的字符串是有效的罗马数字，则任务是通过重新排列给定字符串的字符来打印可能的最高罗马数字。如果发现不可能，则打印“无效”。
例子：
Input: S = “MCMIV”
Output: MMCVI
Explanation: The given string S is valid Roman numeral. Rearranging characters to obtain the string “MMCVI” generates the highest roman numeral possible using given set of characters.
Input: S = “XCYVM”
Output: Invalid
方法：想法是使用第二个元素进行排序。请按照以下步骤解决问题：
- 遍历字符串的字符。
- 检查字符串包含除{ ‘I’, ‘V’, ‘X’, ‘L’, ‘C’, ‘D’, ‘M’ }以外的任何字符。如果发现为真，则打印“无效”。
- 否则，排序相对于字符串中的字符的十进制值按降序排列的字符串。
- 现在将罗马值转换为等效的十进制值并将其存储在一个变量中，比如X。
- 现在找到十进制数X的等效罗马值并将其存储在一个变量中，比如Y。
- 现在打印字符串S如果发现S等于Y 。否则打印“无效”。
下面是上述方法的实现：
C++
// C++ implementation of the above approach
#include
using namespace std;
// Function to find decimal
// value of a roman character
int charVal(char c)
{
if (c == 'I')
return 1;
if (c == 'V')
return 5;
if (c == 'X')
return 10;
if (c == 'L')
return 50;
if (c == 'C')
return 100;
if (c == 'D')
return 500;
if (c == 'M')
return 1000;
return -1;
}
// Function to convert string representing
// roman number to equivalent decimal value
int romanToDec(string S)
{
// Stores the decimal value
// of the string S
int res = 0;
// Stores the size of the S
int n = S.size();
// Update res
res = charVal(S[n - 1]);
// Traverse the string
for (int i = n - 2; i >= 0; i--) {
if (charVal(S[i]) < charVal(S[i + 1]))
res -= charVal(S[i]);
else
res += charVal(S[i]);
}
// Return res
return res;
}
// Function to convert decimal
// to equivalent roman numeral
string DecToRoman(int number)
{
// Stores the string
string res = "";
// Stores all the digit values of a roman digit
int num[] = { 1, 4, 5, 9, 10, 40, 50,
90, 100, 400, 500, 900, 1000 };
string sym[]
= { "I", "IV", "V", "IX", "X", "XL", "L",
"XC", "C", "CD", "D", "CM", "M" };
int i = 12;
// Iterate while number
// is greater than 0
while (number > 0) {
int div = number / num[i];
number = number % num[i];
while (div--) {
res += sym[i];
}
i--;
}
// Return res
return res;
}
// Functur to sort the string
// in descending order of values
// assigned to characters
bool compare(char x, char y)
{
// Return character with
// highest decimal value
int val_x = charVal(x);
int val_y = charVal(y);
return (val_x > val_y);
}
// Function to find largest roman
// value possible by rearranging
// the characters of the string
string findLargest(string S)
{
// Stores all roman characters
set st = { 'I', 'V', 'X', 'L', 'C', 'D', 'M' };
// Traverse the string
for (auto x : S) {
// If X is not found
if (st.find(x) == st.end())
return "Invalid";
}
sort(S.begin(), S.end(), compare);
// Stores the decimal value
// of the roman number
int N = romanToDec(S);
// Find the roman value equivalent
// to the decimal value of N
string R = DecToRoman(N);
if (S != R)
return "Invalid";
// Return result
return S;
}
// Driver Code
int main()
{
string S = "MCMIV";
cout << findLargest(S);
} Java
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG{
// Function to find decimal
// value of a roman character
static int charVal(char c)
{
if (c == 'I')
return 1;
if (c == 'V')
return 5;
if (c == 'X')
return 10;
if (c == 'L')
return 50;
if (c == 'C')
return 100;
if (c == 'D')
return 500;
if (c == 'M')
return 1000;
return -1;
}
// Function to convert string representing
// roman number to equivalent decimal value
static int romanToDec(String S)
{
// Stores the decimal value
// of the string S
int res = 0;
// Stores the size of the S
int n = S.length();
// Update res
res = charVal(S.charAt(n - 1));
// Traverse the string
for(int i = n - 2; i >= 0; i--)
{
if (charVal(S.charAt(i)) <
charVal(S.charAt(i + 1)))
res -= charVal(S.charAt(i));
else
res += charVal(S.charAt(i));
}
// Return res
return res;
}
// Function to convert decimal
// to equivalent roman numeral
static String DecToRoman(int number)
{
// Stores the string
String res = "";
// Stores all the digit values of a roman digit
int num[] = { 1, 4, 5, 9, 10, 40, 50,
90, 100, 400, 500, 900, 1000 };
String sym[] = { "I", "IV", "V", "IX",
"X", "XL", "L", "XC",
"C", "CD", "D", "CM", "M" };
int i = 12;
// Iterate while number
// is greater than 0
while (number > 0)
{
int div = number / num[i];
number = number % num[i];
while (div-- > 0)
{
res += sym[i];
}
i--;
}
// Return res
return res;
}
// Function to find largest roman
// value possible by rearranging
// the characters of the string
static String findLargest(String S)
{
char arr[] = { 'I', 'V', 'X', 'L', 'C', 'D', 'M' };
// Stores all roman characters
Set st = new HashSet<>();
for(char x : arr)
st.add(x);
Character s[] = new Character[S.length()];
for(int i = 0; i < S.length(); i++)
s[i] = S.charAt(i);
// Traverse the string
for(char x : s)
{
// If X is not found
if (!st.contains(x))
return "Invalid";
}
Arrays.sort(s, new Comparator()
{
@Override
public int compare(Character x, Character y)
{
return charVal(y) - charVal(x);
}
});
String ss = "";
for(char ch : s)
ss += ch;
// Stores the decimal value
// of the roman number
int N = romanToDec(ss);
// Find the roman value equivalent
// to the decimal value of N
String R = DecToRoman(N);
if (!ss.equals(R))
return "Invalid";
// Return result
return ss;
}
// Driver Code
public static void main(String[] args)
{
// Input
String S = "MCMIV";
int N = S.length();
System.out.println(findLargest(S));
}
}
// This code is contributed by Kingash 输出：
MMCVI时间复杂度： O(N*log(N))
辅助空间： O(1)
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