给定一个数组，一个整数的arr []和一个整数K。任务是按照给定数组的元素以K为模的升序对它们进行排序。如果两个数字的余数相同，则较小的数字应排在第一位。
例子：
Input: arr[] = {10, 3, 2, 6, 12}, K = 4
Output: 12 2 6 10 3
{12, 2, 6, 10, 3} is the required sorted order as the modulo
of these elements with K = 4 is {0, 2, 2, 2, 3}.
Input: arr[] = {3, 4, 5, 10, 11, 1}, K = 3
Output: 3 1 4 10 5 11
方法：
- 创建K个空向量。
- 从左到右遍历数组并更新向量，以使第i个向量包含除以K时将i作为余数的元素。
- 对所有向量分别进行排序，因为所有使用K赋予相同模值的元素都必须按升序进行排序。
- 现在，从第一个向量到最后一个向量，再从向量中的左到右，将按所需的排序顺序提供元素。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Utility function to print the
// contents of an array
void printArr(int arr[], int n)
{
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
}
// Function to sort the array elements
// based on their modulo with K
void sortWithRemainder(int arr[], int n, int k)
{
// Create K empty vectors
vector v[k];
// Update the vectors such that v[i]
// will contain all the elements
// that give remainder as i
// when divided by k
for (int i = 0; i < n; i++) {
v[arr[i] % k].push_back(arr[i]);
}
// Sorting all the vectors separately
for (int i = 0; i < k; i++)
sort(v[i].begin(), v[i].end());
// Replacing the elements in arr[] with
// the required modulo sorted elements
int j = 0;
for (int i = 0; i < k; i++) {
// Add all the elements of the
// current vector to the array
for (vector::iterator it = v[i].begin();
it != v[i].end(); it++) {
arr[j] = *it;
j++;
}
}
// Print the sorted array
printArr(arr, n);
}
// Driver code
int main()
{
int arr[] = { 10, 7, 2, 6, 12, 3, 33, 46 };
int n = sizeof(arr) / sizeof(arr[0]);
int k = 4;
sortWithRemainder(arr, n, k);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG{
// Utility function to print the
// contents of an array
static void printArr(int[] arr, int n)
{
for(int i = 0; i < n; i++)
System.out.print(arr[i] + " ");
}
// Function to sort the array elements
// based on their modulo with K
static void sortWithRemainder(int[] arr,
int n, int k)
{
// Create K empty vectors
ArrayList<
ArrayList> v = new ArrayList<
ArrayList>(k);
for(int i = 0; i < k; i++)
v.add(new ArrayList());
// Update the vectors such that v[i]
// will contain all the elements
// that give remainder as i
// when divided by k
for(int i = 0; i < n; i++)
{
int t = arr[i] % k;
v.get(t).add(arr[i]);
}
// Sorting all the vectors separately
for(int i = 0; i < k; i++)
{
Collections.sort(v.get(i));
}
// Replacing the elements in
// arr[] with the required
// modulo sorted elements
int j = 0;
for(int i = 0; i < k; i++)
{
// Add all the elements of the
// current vector to the array
for(int x : v.get(i))
{
arr[j] = x;
j++;
}
}
// Print the sorted array
printArr(arr, n);
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 10, 7, 2, 6,
12, 3, 33, 46 };
int n = arr.length;
int k = 4;
sortWithRemainder(arr, n, k);
}
}
// This code is contributed by grand_master Python3
# Python3 implementation of the approach
# Utility function to prthe
# contents of an array
def printArr(arr, n):
for i in range(n):
print(arr[i], end = ' ')
# Function to sort the array elements
# based on their modulo with K
def sortWithRemainder(arr, n, k):
# Create K empty vectors
v = [[] for i in range(k)]
# Update the vectors such that v[i]
# will contain all the elements
# that give remainder as i
# when divided by k
for i in range(n):
v[arr[i] % k].append(arr[i])
# Sorting all the vectors separately
for i in range(k):
v[i].sort()
# Replacing the elements in arr[] with
# the required modulo sorted elements
j = 0
for i in range(k):
# Add all the elements of the
# current vector to the array
for it in v[i]:
arr[j] = it
j += 1
# Print the sorted array
printArr(arr, n)
# Driver code
if __name__=='__main__':
arr = [ 10, 7, 2, 6, 12, 3, 33, 46 ]
n = len(arr)
k = 4
sortWithRemainder(arr, n, k)
# This code is contributed by pratham76C#
// C# implementation of the
// above approach
using System;
using System.Collections;
class GFG{
// Utility function to print the
// contents of an array
static void printArr(int []arr,
int n)
{
for (int i = 0; i < n; i++)
Console.Write(arr[i] + " ");
}
// Function to sort the array elements
// based on their modulo with K
static void sortWithRemainder(int []arr,
int n, int k)
{
// Create K empty vectors
ArrayList []v = new ArrayList[k];
for(int i = 0; i < k; i++)
{
v[i] = new ArrayList();
}
// Update the vectors such that v[i]
// will contain all the elements
// that give remainder as i
// when divided by k
for (int i = 0; i < n; i++)
{
v[arr[i] % k].Add(arr[i]);
}
// Sorting all the vectors separately
for (int i = 0; i < k; i++)
{
v[i].Sort();
}
// Replacing the elements in
// arr[] with the required
// modulo sorted elements
int j = 0;
for (int i = 0; i < k; i++)
{
// Add all the elements of the
// current vector to the array
foreach(int x in v[i])
{
arr[j] = x;
j++;
}
}
// Print the sorted array
printArr(arr, n);
}
// Driver Code
public static void Main(string[] args)
{
int []arr = {10, 7, 2, 6,
12, 3, 33, 46};
int n = arr.Length;
int k = 4;
sortWithRemainder(arr, n, k);
}
}
// This code is contributed by rutvik_56输出：
12 33 2 6 10 46 3 7时间复杂度： O(nlogn)