通过用给定的替代数字替换一段数字来最大化给定的数字

给定许多N位数字。我们还得到了10 个数字，它们代表从0到9 的所有一位数字的替代数字。我们可以用给定的替代数字替换N 中的任何数字，但我们只允许替换任何连续的数字段一次，任务是替换任何连续的数字段，使得获得的数字是最大的所有可能的替代品。

例子：

Input: n = 1337, a[] = {0, 1, 2, 5, 4, 6, 6, 3, 1, 9} 
Output: 1557 
1 can be replaced with 1 as a[1] = 1 (No effect) 
3 can be replaced with 5 
7 can be replaced with 3 (isn’t required if the number needs to be maximized)
Input: number = 11111, a[] = {0, 5, 2, 5, 4, 6, 6, 3, 1, 9} 
Output: 55555  

处理方法：由于我们需要得到最大的数，所以从左边开始迭代，找到比当前数大的那个数，不断替换后面的数，直到这个数不小于这个数。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the maximized number
string get_maximum(string s, int a[])
{
    int n = s.size();
 
    // Iterate till the end of the string
    for (int i = 0; i < n; i++) {
 
        // Check if it is greater or not
        if (s[i] - '0' < a[s[i] - '0']) {
            int j = i;
 
            // Replace with the alternate till smaller
            while (j < n && (s[j] - '0' <= a[s[j] - '0'])) {
                s[j] = '0' + a[s[j] - '0'];
                j++;
            }
 
            return s;
        }
    }
 
    // Return original s in case
    // no change took place
    return s;
}
 
// Driver Code
int main()
{
    string s = "1337";
    int a[] = { 0, 1, 2, 5, 4, 6, 6, 3, 1, 9 };
    cout << get_maximum(s, a);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
 
// Function to return the maximized number
static String get_maximum(char[] s, int a[])
{
    int n = s.length;
 
    // Iterate till the end of the string
    for (int i = 0; i < n; i++)
    {
 
        // Check if it is greater or not
        if (s[i] - '0' < a[s[i] - '0'])
        {
            int j = i;
 
            // Replace with the alternate till smaller
            while (j < n && (s[j] - '0' <= a[s[j] - '0']))
            {
                s[j] = (char) ('0' + a[s[j] - '0']);
                j++;
            }
 
            return String.valueOf(s);
        }
    }
 
    // Return original s in case
    // no change took place
    return String.valueOf(s);
}
 
// Driver Code
public static void main(String[] args)
{
    String s = "1337";
    int a[] = { 0, 1, 2, 5, 4, 6, 6, 3, 1, 9 };
    System.out.println(get_maximum(s.toCharArray(), a));
}
}
 
/* This code contributed by PrinciRaj1992 */

Python3

# Python3 implementation of the approach
 
# Function to return the maximized number
def get_maximum(s, a) :
    s = list(s)
    n = len(s)
     
    # Iterate till the end of the string
    for i in range(n) :
         
        # Check if it is greater or not
        if (ord(s[i]) - ord('0') < a[ord(s[i]) - ord('0')]) :
            j = i
             
            # Replace with the alternate till smaller
            while (j < n and (ord(s[j]) - ord('0') <=
                            a[ord(s[j]) - ord('0')])) :
                s[j] = chr(ord('0') + a[ord(s[j]) - ord('0')])
                j += 1
             
            return "".join(s);
     
    # Return original s in case
    # no change took place
    return s
 
 
# Driver Code
if __name__ == "__main__" :
     
    s = "1337"
    a = [ 0, 1, 2, 5, 4, 6, 6, 3, 1, 9 ]
    print(get_maximum(s, a))
 
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
// Function to return the maximized number
static String get_maximum(char[] s, int[] a)
{
    int n = s.Length;
 
    // Iterate till the end of the string
    for (int i = 0; i < n; i++)
    {
 
        // Check if it is greater or not
        if (s[i] - '0' < a[s[i] - '0'])
        {
            int j = i;
 
            // Replace with the alternate till smaller
            while (j < n && (s[j] - '0' <= a[s[j] - '0']))
            {
                s[j] = (char) ('0' + a[s[j] - '0']);
                j++;
            }
 
            return String.Join("",s);
        }
    }
 
    // Return original s in case
    // no change took place
    return String.Join("",s);
}
 
// Driver Code
public static void Main(String[] args)
{
    String s = "1337";
    int[] a = { 0, 1, 2, 5, 4, 6, 6, 3, 1, 9 };
    Console.WriteLine(get_maximum(s.ToCharArray(), a));
}
}
 
// This code contributed by Rajput-Ji

Javascript

    // Javascript implementation of the approach
     
    // Function to return the maximized number
    function get_maximum(s, a)
    {
        let n = s.length;
 
        // Iterate till the end of the string
        for (let i = 0; i < n; i++)
        {
 
            // Check if it is greater or not
            if (s[i].charCodeAt() - '0'.charCodeAt()
            < a[s[i].charCodeAt() - '0'.charCodeAt()])
            {
                let j = i;
 
                // Replace with the alternate till smaller
                while (j < n && (s[j].charCodeAt() - '0'.charCodeAt()
                <= a[s[j].charCodeAt() - '0'.charCodeAt()]))
                {
                    s[j] = String.fromCharCode('0'.charCodeAt()
                    + a[s[j].charCodeAt() - '0'.charCodeAt()]);
                    j++;
                }
 
                return s.join("");
            }
        }
 
        // Return original s in case
        // no change took place
        return s.join("");
    }
     
    let s = "1337";
    let a = [ 0, 1, 2, 5, 4, 6, 6, 3, 1, 9 ];
    document.write(get_maximum(s.split(''), a));
 

输出：

1557

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