找到最大的偶数整数，它是 S 的非空子串

给定一个大小为N的字符串S ，表示一个大整数。任务是找到最大的偶数整数，它是一个非空子字符串 的S。如果不能生成偶数，则返回一个空字符串。

例子：

Input: S = “4206”
Output: “4206”
Explanation: “4206” is already an even number.

Input: S = “23”
Output: “2”
Explanation: “The only non-empty substrings are “2”, “3”, and “23”. “2” is the only even number.

Input: S = “17”
Output: “”
Explanation: There is no even valued substring in the given string

方法：可以通过从右边找到第一个偶数字符来解决该任务，假设它在索引' idx '处找到。
生成的最大偶数非空子字符串将是[0, idx]范围内 S 的子字符串。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the largest even valued
// substring
void get(string& s)
{
    int N = s.length();
    int idx = -1;
 
    // Finding the rightmost even character
    for (int i = N - 1; i >= 0; i--) {
        if ((s[i] - '0') % 2 == 0) {
            idx = i;
            break;
        }
    }
 
    if (idx == -1)
        cout << "";
    else
        cout << s.substr(0, idx + 1);
}
 
// Driver Code
int main()
{
    string S = "4206";
    get(S);
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG {
 
  // Function to find the largest even valued
  // substring
  static void get(String s)
  {
    int N = s.length();
    int idx = -1;
 
    // Finding the rightmost even character
    for (int i = N - 1; i >= 0; i--) {
      if ((s.charAt(i) - '0') % 2 == 0) {
        idx = i;
        break;
      }
    }
 
    if (idx == -1)
      System.out.print("");
    else
      System.out.print(s.substring(0, idx + 1));
  }
 
  // Driver Code
  public static void main (String[] args) {
    String S = "4206";
    get(S);
  }
}
 
// This code is contributed by hrithikgarg03188.

Python3

# Python code for the above approach
 
# Function to find the largest even valued
# substring
def get(s):
    N = len(s);
    idx = -1;
 
    # Finding the rightmost even character
    for i in range(N - 1, 0, -1):
        if ((ord(s[i]) - ord('0')) % 2 == 0):
            idx = i;
            break;
 
    if (idx == -1):
        print("");
    else:
        print(s[0: idx + 1]);
 
# Driver Code
S = "4206";
get(S);
 
# This code is contributed by gfgking

C#

// C# program for the above approach
using System;
class GFG {
 
  // Function to find the largest even valued
  // substring
  static void get(string s)
  {
    int N = s.Length;
    int idx = -1;
 
    // Finding the rightmost even character
    for (int i = N - 1; i >= 0; i--) {
      if ((s[i] - '0') % 2 == 0) {
        idx = i;
        break;
      }
    }
 
    if (idx == -1)
      Console.Write("");
    else
      Console.Write(s.Substring(0, idx + 1));
  }
 
  // Driver Code
  public static void Main () {
    string S = "4206";
    get(S);
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

    // JavaScript code for the above approach
 
    // Function to find the largest even valued
    // substring
    function get(s)
    {
        let N = s.length;
        let idx = -1;
 
        // Finding the rightmost even character
        for (let i = N - 1; i >= 0; i--)
        {
            if ((s[i].charCodeAt(0) - '0'.charCodeAt(0)) % 2 == 0)
            {
                idx = i;
                break;
            }
        }
 
        if (idx == -1)
            document.write("");
        else
            document.write(s.slice(0, idx + 1));
    }
 
    // Driver Code
    let S = "4206";
    get(S);
 
   // This code is contributed by Potta Lokesh

输出

4206

时间复杂度： O(N)
辅助空间： O(1)