数组中回收对的数量
给定一个整数数组 arr[]，找出数组中回收对的数量。两个数字的循环对 {a, b} 具有以下属性：
- A 应该小于 B。
- 位数应该相同。
- 通过在一个方向上旋转 A 任意次数，我们应该得到 B。
例子：
Input : arr[] = {32, 42, 13, 23, 9, 5, 31}
Output : 2
Explanation : Since there are two pairs {13, 31} and {23, 32}.
By rotating 13 for first time, output is 31 and by rotating 23 once output is 32.
Both of these pairs satisfy our criteria.
Input : arr[] = {1212, 2121}
Output : 1
Explanation : Since there are two pairs {1212, 2121}. By rotating 1212
for first time, output is 2121. This pair satisfies our criteria.
Note that if rotation id done further, rotating 1212 again output is 1212
which is given number and 2121 which has been already counted.
So discard both of these results.
下面是解决上述问题的分步算法：
- 对数组进行排序。
- 创建一个大小为 n 的新数组“temp”，其中 n 是原始数组的长度。
- 通过将唯一值复制到新数组“temp”来从数组中删除重复项。
- 找到从原始数组复制的元素数量，并将这个数字作为数组的大小。
- 创建一个 HashSet 以仅存储当前数字的唯一旋转。
- 用 value = 0 初始化一个计数器。
- 遍历“temp”并为每个数字执行以下步骤 -
- 找出位数。让它成为'd1'。
- 将数字旋转 d-1 次，并将每次旋转形成的每个数字存储在 HashSet 中。
- 如果在 HashSet 中找到形成的数字，则忽略它。
- 对于每个旋转的数字，对它在数组的其余部分中的存在进行二进制搜索。
- 如果存在，则递增计数器。
C++
// C++ code for Recycled Pairs in array.
#include
using namespace std;
// Function to find recycled pairs
int recycledPairs(int a[], int n)
{
int count = 0;
// Sorting array
sort(a, a + n);
// Removing duplicates by creating new array temp.
int temp[n];
memset(temp, -1, n);
int j = 0;
for (int i = 0; i < n - 1; i++)
if (a[i] != a[i + 1])
temp[j++] = a[i];
temp[j++] = a[n - 1];
int size = n;
// Finding number of locations in temp
// which are occupied from copying.
for (int i = n - 1; i >= 0; i--)
if (temp[i] != -1)
{
size = i;
break;
}
// Hashset to store new Rotations
seths;
for (int i = 0; i < size + 1; i++)
{
// Clearing hashset for each number in temp.
hs.clear();
int x = temp[i];
// Finding number of digits of taken number
int d1 = (int)log10(temp[i]) + 1;
int f = (int)pow(10, d1 - 1);
for (j = 1; j <= d1 - 1; j++)
{
// Remainder
int r = x % 10;
// Quotient
int q = x / 10;
// Forming new number by rotating.
x = r * f + q;
// Number of digits of newly formed rotated number
// to avoid duplicate numbers.
int d2 = (int)log10(x) + 1;
set::iterator it = hs.find(x);
// Inserting formed rotated number to set s
if (it == hs.end())
{
hs.insert(x);
// Checking for number of digits of new number.
if ((d1 == d2))
{
// Searching for the formed element in rest of array.
int position = lower_bound(temp + i,
temp + size + 1 , x)-(temp + i + 1);
// If position found
if(position >= 0)
{
// Increment counter.
count++;
}
}
}
}
}
// Return counter
return count;
}
// Driver function
int main()
{
int a[] = { 32, 42, 13, 23, 9, 5, 31 };
int n = sizeof(a)/sizeof(a[0]);
int result = recycledPairs(a,n);
cout << (result);
return 0;
}
// This code is contributed by Rajput-Ji Java
// Java code for Recycled Pairs in array.
import java.util.*;
class GFG {
// Function to find recycled pairs
static int recycledPairs(int[] a)
{
int count = 0;
// Sorting array
Arrays.sort(a);
int n = a.length;
// Removing duplicates by creating new array temp.
int[] temp = new int[n];
Arrays.fill(temp, -1);
int j = 0;
for (int i = 0; i < n - 1; i++)
if (a[i] != a[i + 1])
temp[j++] = a[i];
temp[j++] = a[n - 1];
int size = n;
// Finding number of locations in temp which are occupied from copying.
for (int i = n - 1; i >= 0; i--)
if (temp[i] != -1) {
size = i;
break;
}
// Hashset to store new Rotations
HashSet hs = new HashSet();
for (int i = 0; i < size + 1; i++) {
// Clearing hashset for each number in temp.
hs.clear();
int x = temp[i];
// Finding number of digits of taken number
int d1 = (int)Math.log10(temp[i]) + 1;
int f = (int)Math.pow(10, d1 - 1);
for (j = 1; j <= d1 - 1; j++) {
// Remainder
int r = x % 10;
// Quotient
int q = x / 10;
// Forming new number by rotating.
x = r * f + q;
// Number of digits of newly formed rotated number
// to avoid duplicate numbers.
int d2 = (int)Math.log10(x) + 1;
// Inserting formed rotated number to set s
if (!hs.contains(x)) {
hs.add(x);
// Checking for number of digits of new number.
if ((d1 == d2))
{
// Searching for the formed element in rest of array.
int position = Arrays.binarySearch(temp, i + 1, size + 1, x);
// If position found
if(position >= 0)
{
// Increment counter.
count++;
}
}
}
}
}
// Return counter
return count;
}
// Driver function
public static void main(String[] args)
{
int a[] = { 32, 42, 13, 23, 9, 5, 31 };
int result = recycledPairs(a);
System.out.println(result);
}
} C#
// C# code for Recycled Pairs in array.
using System;
using System.Collections.Generic;
class GFG
{
// Function to find recycled pairs
static int recycledPairs(int[] a)
{
int count = 0;
// Sorting array
Array.Sort(a);
int n = a.Length;
// Removing duplicates by
// creating new array temp.
int[] temp = new int[n];
for (int i = 0; i < n; i++)
temp[i] = -1;
int j = 0;
for (int i = 0; i < n - 1; i++)
if (a[i] != a[i + 1])
temp[j++] = a[i];
temp[j++] = a[n - 1];
int size = n;
// Finding number of locations in temp
// which are occupied from copying.
for (int i = n - 1; i >= 0; i--)
if (temp[i] != -1)
{
size = i;
break;
}
// Hashset to store new Rotations
HashSet hs = new HashSet();
for (int i = 0; i < size + 1; i++)
{
// Clearing hashset for each number in temp.
hs.Clear();
int x = temp[i];
// Finding number of digits of taken number
int d1 = (int)Math.Log10(temp[i]) + 1;
int f = (int)Math.Pow(10, d1 - 1);
for (j = 1; j <= d1 - 1; j++)
{
// Remainder
int r = x % 10;
// Quotient
int q = x / 10;
// Forming new number by rotating.
x = r * f + q;
// Number of digits of newly formed rotated number
// to avoid duplicate numbers.
int d2 = (int)Math.Log10(x) + 1;
// Inserting formed rotated number to set s
if (!hs.Contains(x))
{
hs.Add(x);
// Checking for number of digits of new number.
if ((d1 == d2))
{
// Searching for the formed element in rest of array.
int position = Array.BinarySearch(temp, i + 1,
size - i, x);
// If position found
if(position >= 0)
{
// Increment counter.
count++;
}
}
}
}
}
// Return counter
return count;
}
// Driver Code
public static void Main(String[] args)
{
int []a = { 32, 42, 13, 23, 9, 5, 31 };
int result = recycledPairs(a);
Console.WriteLine(result);
}
}
// This code is contributed by 29AjayKumar 输出：
2
时间复杂度：O(n*log(n))。
注意：对于任何给定的整数，形成新数字的最大旋转次数是固定的，即 (no_of_digits-1)。因此，此操作是常数时间，即 O(1)。