从给定的数字数组中查找所有不同的三位数字

给定一个包含digits[]的数组，其中每个元素都是一个数字整数。该数组可能包含重复项。任务是找到所有满足给定要求的唯一整数：

• 整数由任意顺序的数字的三个元素的串联组成。
• 整数没有前导零。

例子：

Input: digits[] = {2, 1, 3, 0}
Output: {102, 103, 120, 123, 130, 132, 201, 203, 210, 213, 230, 231, 301, 302, 310, 312, 320, 321}
Explanation: The above are the three digit numbers formed.  

Input: digits[] = {3, 7, 5}
Output: [357, 375, 537, 573, 735, 753 ]

方法：这个问题可以通过使用频率图来解决。查找给定数字数组中所有元素的计数。请按照以下步骤解决给定的问题。

• 检查100 到 999之间的所有数字是否可以由数字向量中存在的数字组成。
• 使用2张地图相同。如果可以制作数字，则将其添加到答案中。
• 最后，返回答案。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find all the unique
// 3 digit number that can be formed
// from the given digits
vector find3DigitNumbers(vector&
                                  digits)
{
    // Generating frequency map
    // of the given digits
    vector count(10, 0);
    for (auto& d : digits)
        count[d]++;
 
    vector res;
 
    for (int num = 100; num < 999; num++) {
 
        // Generating frequency map
        // of the current number
        vector currCount(10, 0);
        int temp = num;
 
        while (temp) {
            currCount[temp % 10]++;
            temp /= 10;
        }
 
        // Checking if the number
        // can be generated or not
        bool flag = true;
 
        for (int i = 0; i < 10; i++) {
            if (currCount[i] > count[i]) {
                flag = false;
                break;
            }
        }
 
        if (flag) {
            res.push_back(num);
        }
    }
    return res;
}
 
// Function to print answer
void printAnswer(vector& v1)
{
    for (int i = 0; i < v1.size(); i++) {
        cout << v1[i] << " ";
    }
    cout << endl;
}
 
// Driver code
int main()
{
    vector v1 = { 2, 1, 3, 0 };
 
    // Function Call
    vector ans = find3DigitNumbers(v1);
 
    // Printing answer
    printAnswer(ans);
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG {
 
  // Function to find all the unique
  // 3 digit number that can be formed
  // from the given digits
  static void find3DigitNumbers(int digits[], List res)
  {
    // Generating frequency map
    // of the given digits
    int count[] = new int[10];;
    for (int i = 0; i < digits.length; i++)
      count[digits[i]]++;
 
    for (int num = 100; num < 999; num++) {
 
      // Generating frequency map
      // of the current number
      int currCount[] = new int[10];
      int temp = num;
 
      while (temp > 0) {
        currCount[temp % 10]++;
        temp /= 10;
      }
 
      // Checking if the number
      // can be generated or not
      Boolean flag = true;
 
      for (int i = 0; i < 10; i++) {
        if (currCount[i] > count[i]) {
          flag = false;
          break;
        }
      }
 
      if (flag == true) {
        res.add(num);
      }
    }
  }
 
  // Function to print answer
  static void printAnswer(List res)
  {
    for(int i = 0;  i < res.size(); i++)
      System.out.print(res.get(i) + " ");
  }
 
  // Driver code
  public static void main (String[] args)
  {
 
    int arr[] = { 2, 1, 3, 0 };
 
    List ans=new ArrayList();
    // Function Call
    find3DigitNumbers(arr, ans);
 
    // Printing answer
    printAnswer(ans);
  }
}
 
// This code is contributed by hrithikgarg03188.

Python3

# Python code for the above approach
 
# Function to find all the unique
# 3 digit number that can be formed
# from the given digits
def find3DigitNumbers(digits):
   
    # Generating frequency map
    # of the given digits
    count = [0] * 10
    for d in digits:
        count[d] += 1
 
    res = []
 
    for num in range(100, 999):
 
        # Generating frequency map
        # of the current number
        currCount = [0] * 10
        temp = num
 
        while (temp):
            currCount[temp % 10] += 1
            temp = temp // 10
 
        # Checking if the number
        # can be generated or not
        flag = True
 
        for i in range(10):
            if (currCount[i] > count[i]):
                flag = False
                break
 
        if (flag):
            res.append(num)
 
    return res
 
# Function to print answer
def printAnswer(v1):
    for i in range(len(v1)):
        print(v1[i], end=" ")
 
    print('')
 
# Driver code
v1 = [2, 1, 3, 0]
 
# Function Call
ans = find3DigitNumbers(v1)
 
# Printing answer
printAnswer(ans)
 
 
# This code is contributed by Saurabh Jaiswal

C#

// C# program for the above approach
using System;
using System.Collections;
class GFG {
 
  // Function to find all the unique
  // 3 digit number that can be formed
  // from the given digits
  static void find3DigitNumbers(int []digits, ArrayList res)
  {
    // Generating frequency map
    // of the given digits
    int []count = new int[10];;
    for (int i = 0; i < digits.Length; i++)
      count[digits[i]]++;
 
    for (int num = 100; num < 999; num++) {
 
      // Generating frequency map
      // of the current number
      int []currCount = new int[10];
      int temp = num;
 
      while (temp > 0) {
        currCount[temp % 10]++;
        temp /= 10;
      }
 
      // Checking if the number
      // can be generated or not
      bool flag = true;
 
      for (int i = 0; i < 10; i++) {
        if (currCount[i] > count[i]) {
          flag = false;
          break;
        }
      }
 
      if (flag == true) {
        res.Add(num);
      }
    }
  }
 
  // Function to print answer
  static void printAnswer(ArrayList res)
  {
    for(int i = 0;  i < res.Count; i++)
      Console.Write(res[i] + " ");
  }
 
  // Driver code
  public static void Main ()
  {
 
    int []arr = { 2, 1, 3, 0 };
 
    ArrayList ans=new ArrayList();
    // Function Call
    find3DigitNumbers(arr, ans);
 
    // Printing answer
    printAnswer(ans);
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

       // JavaScript code for the above approach
 
       // Function to find all the unique
       // 3 digit number that can be formed
       // from the given digits
       function find3DigitNumbers(
           digits) {
           // Generating frequency map
           // of the given digits
           let count = new Array(10).fill(0);
           for (let d of digits)
               count[d]++;
 
           let res = [];
 
           for (let num = 100; num < 999; num++) {
 
               // Generating frequency map
               // of the current number
               let currCount = new Array(10).fill(0);
               let temp = num;
 
               while (temp) {
                   currCount[temp % 10]++;
                   temp = Math.floor(temp / 10);
               }
 
               // Checking if the number
               // can be generated or not
               let flag = true;
 
               for (let i = 0; i < 10; i++) {
                   if (currCount[i] > count[i]) {
                       flag = false;
                       break;
                   }
               }
 
               if (flag) {
                   res.push(num);
               }
           }
           return res;
       }
 
       // Function to print answer
       function printAnswer(v1) {
           for (let i = 0; i < v1.length; i++) {
               document.write(v1[i] + " ");
           }
           document.write('<br>')
       }
 
       // Driver code
 
       let v1 = [2, 1, 3, 0];
 
       // Function Call
       let ans = find3DigitNumbers(v1);
 
       // Printing answer
       printAnswer(ans);
 
 
      // This code is contributed by Potta Lokesh
   

输出

102 103 120 123 130 132 201 203 210 213 230 231 301 302 310 312 320 321 

时间复杂度： O(1000)
辅助空间： O(N)