📜  通过交换数字最近的更大的数字

📅  最后修改于: 2021-09-08 15:08:39             🧑  作者: Mango

给定两个整数AB 。任务是通过交换A的数字来找到与B最接近的更大值。如果没有这样的排列可能,则打印 -1。
例子:

先决条件:字符串的所有排列
方法:

  • 使用 Integer.MAX_VALUE 设置 min1 的最小值
  • 用上面提到的排列方法交换A的数字。
  • 检查 A 的排列是否小于 min1。如果小于则将 min1 更新为 A。
  • 对 A 的所有排列重复此操作并找到最小的较大值

下面是上述方法的实现:

C++
// C++ program to find nearest greater value
#include 
using namespace std;
 
int min1 = INT_MAX;
int _count = 0;
 
// Find all the possible permutation of Value A.
int permutation(string str1, int i, int n, int p)
{
    if (i == n)
    {
        // Convert into Integer
        int q = stoi(str1);
 
        // Find the minimum value of A by interchanging
        // the digit of A and store min1.
        if (q - p > 0 && q < min1)
        {
            min1 = q;
            _count = 1;
        }
    }
    else
    {
        for (int j = i; j <= n; j++)
        {
            // Swap two string character
            swap(str1[i], str1[j]);
            permutation(str1, i + 1, n, p);
            swap(str1[i], str1[j]);
        }
    }
    return min1;
}
 
// Driver code
int main()
{
    int A = 213;
    int B = 111;
 
    // Convert integer value into string to
    // find all the permutation of the number
    string str1 = to_string(A);
    int len = str1.length();
    int h = permutation(str1, 0, len - 1, B);
 
    // count=1 means number greater than B exists
    _count ? cout << h << endl : cout << -1 << endl;
 
    return 0;
}
 
// This code is contributed by
// sanjeev2552


Java
// JAVA program to find nearest greater value
import java.io.*;
import java.util.*;
 
class GFG {
    static int min1 = Integer.MAX_VALUE;
    static int count = 0;
 
    // Find all the possible permutation of Value A.
    public int permutation(String str1, int i, int n, int p)
    {
 
        if (i == n) {
 
            // Convert into Integer
            int q = Integer.parseInt(str1);
 
            // Find the minimum value of A by interchanging
            // the digit of A and store min1.
            if (q - p > 0 && q < min1) {
                min1 = q;
                count = 1;
            }
        }
 
        else {
            for (int j = i; j <= n; j++) {
                str1 = swap(str1, i, j);
                permutation(str1, i + 1, n, p);
                str1 = swap(str1, i, j);
            }
        }
        return min1;
    }
 
    // Swap two string character
    public String swap(String str, int i, int j)
    {
        char ch[] = str.toCharArray();
        char temp = ch[i];
        ch[i] = ch[j];
        ch[j] = temp;
        // Return the string after
        // swapping between two character.
        return String.valueOf(ch);
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        int A = 213;
        int B = 111;
 
        GFG gfg = new GFG();
 
        // Convert integer value into string to
        // find all the permutation of the number
        String str1 = Integer.toString(A);
        int len = str1.length();
        int h = gfg.permutation(str1, 0, len - 1, B);
 
        // count=1 means number greater than B exists
        if (count == 1)
            System.out.println(h);
        else
            System.out.println(-1);
    }
}


Python3
# Python3 program to find nearest greater value
min1 = 10**9
_count = 0
 
# Find all the possible permutation of Value A.
def permutation(str1, i, n, p):
    global min1, _count
    if (i == n):
         
        # Convert into Integer
        str1 = "".join(str1)
        q = int(str1)
 
        # Find the minimum value of A
        # by interchanging the digits
        # of A and store min1.
        if (q - p > 0 and q < min1):
            min1 = q
            _count = 1
    else:
        for j in range(i, n + 1):
             
            # Swap two character)
            str1[i], str1[j] = str1[j], str1[i]
            permutation(str1, i + 1, n, p)
            str1[i], str1[j] = str1[j], str1[i]
 
    return min1
 
# Driver code
A = 213
B = 111
 
# Convert integer value into to
# find all the permutation of the number
str2 = str(A)
str1 = [i for i in str2]
le = len(str1)
 
h = permutation(str1, 0, le - 1, B)
 
# count=1 means number greater than B exists
if _count == 1:
    print(h)
else:
    print(-1)
 
# This code is contributed by
# mohit


C#
// C# program to find nearest greater value
using System;
     
class GFG
{
    static int min1 = int.MaxValue;
    static int count = 0;
 
    // Find all the possible permutation of Value A.
    public int permutation(String str1, int i,
                                 int n, int p)
    {
        if (i == n)
        {
 
            // Convert into Integer
            int q = int.Parse(str1);
 
            // Find the minimum value of A by interchanging
            // the digit of A and store min1.
            if (q - p > 0 && q < min1)
            {
                min1 = q;
                count = 1;
            }
        }
 
        else
        {
            for (int j = i; j <= n; j++)
            {
                str1 = swap(str1, i, j);
                permutation(str1, i + 1, n, p);
                str1 = swap(str1, i, j);
            }
        }
        return min1;
    }
 
    // Swap two string character
    public String swap(String str, int i, int j)
    {
        char []ch = str.ToCharArray();
        char temp = ch[i];
        ch[i] = ch[j];
        ch[j] = temp;
         
        // Return the string after
        // swapping between two character.
        return String.Join("", ch);
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int A = 213;
        int B = 111;
 
        GFG gfg = new GFG();
 
        // Convert integer value into string to
        // find all the permutation of the number
        String str1 = A.ToString();
        int len = str1.Length;
        int h = gfg.permutation(str1, 0, len - 1, B);
 
        // count=1 means number greater than B exists
        if (count == 1)
            Console.WriteLine(h);
        else
            Console.WriteLine(-1);
    }
}
 
// This code is contributed by PrinciRaj1992


Javascript


输出:
123

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