通过重新排列给定正数或负数的数字来获得最大数

给定一个整数N （正数或负数），任务是找到可以使用该数的所有数字组成的最大数。

例子：

Input: -38290367
Output: -20336789
Explanation: As there is need to use all the digits, 0 cannot be the first digit because it becomes redundant at first position.

Input: 1203465
Output: 6543210

方法：数字中的位数范围为 0-9，因此想法是创建一个大小为 10 的哈希数组，并将该数字中出现的每个数字的计数存储在哈希数组中。请按照以下步骤解决问题：

• 然后先检查数字是正数还是负数
• 如果数字是正数，那么如本文所述，从索引9 到 0遍历散列数组并相应地计算数字。
• 但如果数字为负数，则从0-9遍历数组，这样：
  • 可以有前导 0。所以，找到最小的非零数字
  • 在开头插入最小的非零数字，然后按顺序将所有其他数字从 0 到 9添加到最终答案
  • 然后将数字乘以 -1 使其为负数
• 返回结果数字

下面是上述方法的实现

C++

// C++ program to implement the approach
#include 
using namespace std;
 
// Function to print the maximum number
long long printMaxNum(long long num)
{
    // Initialising hash array
    int hash[10] = { 0 };
    long long n = num < 0 ? num * -1 : num;
    long long ans = 0;
    while (n) {
        hash[n % 10]++;
        n = n / 10;
    }
 
    // If positive number
    if (num > 0) {
        for (int i = 9; i >= 0; i--)
            for (int j = 0; j < hash[i]; j++)
                ans = ans * 10 + i;
    }
 
    // If negative number
    else {
 
        // If 0 is present in the number
        if (hash[0] > 0) {
            for (int i = 1; i < 10; i++)
                if (hash[i] > 0) {
                    ans = i;
                    hash[i]--;
                    break;
                }
        }
        for (int i = 0; i < 10; i++)
            for (int j = 0; j < hash[i]; j++)
                ans = ans * 10 + i;
        ans = ans * -1;
    }
    return ans;
}
 
// Driver code
int main()
{
    int N = -38290367;
 
    // Function call
    cout << printMaxNum(N);
    return 0;
}

Java

// Java program to implement the approach
class GFG
{
 
  // Function to print the maximum number
  static long printMaxNum(long num)
  {
 
    // Initialising hash array
    int[] hash = new int[10];
    for (int i = 0; i < 10; i++) {
      hash[i] = 0;
    }
    long n = num < 0 ? num * -1 : num;
    long ans = 0;
    while (n > 0) {
      hash[(int)(n % 10)] += 1;
      n = n / 10;
    }
 
    // If positive number
    if (num > 0) {
      for (int i = 9; i >= 0; i--)
        for (int j = 0; j < hash[i]; j++)
          ans = ans * 10 + i;
    }
 
    // If negative number
    else {
 
      // If 0 is present in the number
      if (hash[0] > 0) {
        for (int i = 1; i < 10; i++)
          if (hash[i] > 0) {
            ans = i;
            hash[i]--;
            break;
          }
      }
      for (int i = 0; i < 10; i++)
        for (int j = 0; j < hash[i]; j++)
          ans = ans * 10 + i;
      ans = ans * -1;
    }
    return ans;
  }
 
  // Driver code
  public static void main(String args[])
  {
    int N = -38290367;
 
    // Function call
    System.out.println(printMaxNum(N));
  }
}
 
// This code is contributed by gfgking

Python3

# Python program to implement the approach
 
# Function to print the maximum number
def printMaxNum(num):
 
    # Initialising hash array
    hash = []
    for i in range(0, 10):
        hash.append(0)
 
    if(num < 0):
        n = num * -1
    else:
        n = num
 
    ans = 0
    while (n != 0):
        hash[int(n % 10)] = hash[int(n % 10)] + 1
        n = n // 10
 
    # If positive number
    if (num > 0):
        for i in range(9, -1, -1):
            for j in range(0, hash[i]):
                ans = ans * 10 + i
 
    # If negative number
    else:
 
        # If 0 is present in the number
        if (hash[0] > 0):
            for i in range(1, 10):
                if (hash[i] > 0):
                    ans = i
                    hash[i] = hash[i]-1
                    break
 
        for i in range(0, 10):
            for j in range(0, hash[i]):
                ans = ans * 10 + i
        ans = ans * -1
 
    return ans
 
# Driver code
N = -38290367
 
# Function call
print(printMaxNum(N))
 
# This code is contributed by Taranpreet

C#

// C# program to implement the approach
using System;
class GFG {
    // Function to print the maximum number
    static long printMaxNum(long num)
    {
        // Initialising hash array
        int[] hash = new int[10];
        for (int i = 0; i < 10; i++) {
            hash[i] = 0;
        }
        long n = num < 0 ? num * -1 : num;
        long ans = 0;
        while (n > 0) {
            hash[n % 10]++;
            n = n / 10;
        }
 
        // If positive number
        if (num > 0) {
            for (int i = 9; i >= 0; i--)
                for (int j = 0; j < hash[i]; j++)
                    ans = ans * 10 + i;
        }
 
        // If negative number
        else {
 
            // If 0 is present in the number
            if (hash[0] > 0) {
                for (int i = 1; i < 10; i++)
                    if (hash[i] > 0) {
                        ans = i;
                        hash[i]--;
                        break;
                    }
            }
            for (int i = 0; i < 10; i++)
                for (int j = 0; j < hash[i]; j++)
                    ans = ans * 10 + i;
            ans = ans * -1;
        }
        return ans;
    }
 
    // Driver code
    public static void Main()
    {
        int N = -38290367;
 
        // Function call
        Console.Write(printMaxNum(N));
    }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

// javascript program to implement the approach
 
  // Function to print the maximum number
  function printMaxNum(num)
  {
 
    // Initialising hash array
    var hash = Array.from({length: 10}, (_, i) => 0);
    for (var i = 0; i < 10; i++) {
      hash[i] = 0;
    }
    var n = num < 0 ? num * -1 : num;
    var ans = 0;
    while (n > 0) {
      hash[parseInt(n % 10)] += 1;
      n = parseInt(n / 10);
    }
 
    // If positive number
    if (num > 0) {
      for (var i = 9; i >= 0; i--)
        for (var j = 0; j < hash[i]; j++)
          ans = ans * 10 + i;
    }
 
    // If negative number
    else {
 
      // If 0 is present in the number
      if (hash[0] > 0) {
        for (var i = 1; i < 10; i++)
          if (hash[i] > 0) {
            ans = i;
            hash[i]--;
            break;
          }
      }
      for (var i = 0; i < 10; i++)
        for (var j = 0; j < hash[i]; j++)
          ans = ans * 10 + i;
      ans = ans * -1;
    }
    return ans;
  }
 
  // Driver code
var N = -38290367;
 
// Function call
document.write(printMaxNum(N));
 
// This code is contributed by shikhasingrajput

输出

-20336789

时间复杂度： O(length(N))
辅助空间： O(1)