给定一个由N 个正整数组成的数组arr[] ，任务是通过将每个数组元素的数字递归相乘来获得一位数所需的步数，以递增的顺序对数组进行排序。如果任意两个数字的步数相同，则先打印较小的数字。
例子：
Input: arr[] = {39, 999, 4, 9876}
Output: 4 9876 39 999
Explanation:
Following are the number of steps required to reduce every array element to 0:
- For arr[0] (= 39): The element 39 will reduce as 39 → 27 → 14 → 4. Therefore, the number of steps required is 3.
- For arr[1] (= 999): The element 999 will reduce as 999 → 729 → 126 → 12 → 2. Therefore, the number of steps required is 4.
- For arr[2] (= 4): The element 4 is already a single-digit number. Therefore, the number of steps required is 0.
- For arr[3] (= 9876): The element 9876 will reduce as 9876 → 3024 → 0. Therefore, the number of steps required is 2.
According to the given criteria the elements in increasing order of count of steps required to reduce them into single-digit number is {4, 9876, 29, 999}
Input: arr[] = {1, 27, 90}
Output: 1 90 27
方法：给定的问题可以通过找到获得一位数所需的步数来解决，方法是将每个数组元素的数字递归相乘，然后使用比较器函数按递增顺序对数组进行排序。请按照以下步骤解决问题：
- 声明一个比较器函数cmp(X, Y) ，它将两个元素作为参数并执行以下步骤：
- 迭代一个循环，直到X变成一位数，并将X的值更新为其数字的乘积。
- 对值Y重复上述步骤。
- 如果X的值小于Y ，则返回true 。否则，返回false 。
- 使用上述比较器函数sort(arr, arr + N, cmp)对给定数组arr[]进行排序。
- 完成上述步骤后，打印数组arr[] 。
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 number of
// steps required to reduce a given
// number to a single-digit number
int countReduction(int num)
{
// Stores the required result
int ans = 0;
// Iterate until a single digit
// number is not obtained
while (num >= 10) {
// Store the number in a
// temporary variable
int temp = num;
num = 1;
// Iterate over all digits and
// store their product in num
while (temp > 0) {
int digit = temp % 10;
temp = temp / 10;
num *= digit;
}
// Increment the answer
// by 1
ans++;
}
// Return the result
return ans;
}
// Comparator function to sort the array
bool compare(int x, int y)
{
// Count number of steps required
// to reduce X and Y into single
// digits integer
int x1 = countReduction(x);
int y1 = countReduction(y);
// Return true
if (x1 < y1)
return true;
return false;
}
// Function to sort the array according
// to the number of steps required to
// reduce a given number into a single
// digit number
void sortArray(int a[], int n)
{
// Sort the array using the
// comparator function
sort(a, a + n, compare);
// Print the array after sorting
for (int i = 0; i < n; i++) {
cout << a[i] << " ";
}
}
// Driver Code
int main()
{
int arr[] = { 39, 999, 4, 9876 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
sortArray(arr, N);
return 0;
} Java
// Java program for the above approach
import java.util.*;
import java.lang.*;
class GFG
{
// Function to find the number of
// steps required to reduce a given
// number to a single-digit number
static int countReduction(int num)
{
// Stores the required result
int ans = 0;
// Iterate until a single digit
// number is not obtained
while (num >= 10)
{
// Store the number in a
// temporary variable
int temp = num;
num = 1;
// Iterate over all digits and
// store their product in num
while (temp > 0) {
int digit = temp % 10;
temp = temp / 10;
num *= digit;
}
// Increment the answer
// by 1
ans++;
}
// Return the result
return ans;
}
// Function to sort the array according
// to the number of steps required to
// reduce a given number into a single
// digit number
static void sortArray(Integer a[], int n)
{
// Sort the array using the
// comparator function
Arrays.sort(a,new Comparator(){
public int compare(Integer x, Integer y)
{
// Count number of steps required
// to reduce X and Y into single
// digits integer
int x1 = countReduction(x);
int y1 = countReduction(y);
// Return true
if (x1 < y1)
return -1;
return 1;
}
});
// Print the array after sorting
for (int i = 0; i < n; i++)
{
System.out.print(a[i] + " ");
}
}
// Driver code
public static void main (String[] args)
{
Integer arr[] = { 39, 999, 4, 9876 };
int N = arr.length;
// Function Call
sortArray(arr, N);
}
}
// This code is contributed by offbeat 输出：
4 9876 39 999时间复杂度： O(N * log(N) * log(X))，其中X是数组中最大的元素，arr[]
辅助空间： O(1)
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