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📜  通过重新排列给定字符串的字符来实现最大的罗马数字

📅  最后修改于: 2021-09-07 02:11:40             🧑  作者: Mango

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:

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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