给定N个非负整数的数组arr [] ，任务是根据它们的倒序对这些整数进行排序
例子：
Input: arr[] = {12, 10, 102, 31, 15}
Output: 10 31 12 15 102
Reversing the numbers:
12 -> 21
10 -> 01
102 -> 201
31 -> 13
15 -> 51
Sorting the reversed numbers: 01 13 21 51 201
Original sorted array: 10 13 12 15 102
Input: arr[] = {12, 10}
Output: 10 12
方法：想法是将每个元素及其反向元素存储在向量对中，然后根据存储的反向元素对向量的所有元素进行排序。最后，按顺序打印元素。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the
// reverse of n
int reversDigits(int num)
{
int rev_num = 0;
while (num > 0) {
rev_num = rev_num * 10 + num % 10;
num = num / 10;
}
return rev_num;
}
// Function to sort the array according to
// the reverse of elements
void sortArr(int arr[], int n)
{
// Vector to store the reverse
// with respective elements
vector > vp;
// Inserting reverse with elements
// in the vector pair
for (int i = 0; i < n; i++) {
vp.push_back(
make_pair(reversDigits(arr[i]),
arr[i]));
}
// Sort the vector, this will sort the pair
// according to the reverse of elements
sort(vp.begin(), vp.end());
// Print the sorted vector content
for (int i = 0; i < vp.size(); i++)
cout << vp[i].second << " ";
}
// Driver code
int main()
{
int arr[] = { 12, 10, 102, 31, 15 };
int n = sizeof(arr) / sizeof(arr[0]);
sortArr(arr, n);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG{
// Function to return the
// reverse of n
static int reversDigits(int num)
{
int rev_num = 0;
while (num > 0)
{
rev_num = rev_num * 10 + num % 10;
num = num / 10;
}
return rev_num;
}
// Function to sort the array according
// to the reverse of elements
static void sortArr(int arr[], int n)
{
// Vector to store the reverse
// with respective elements
ArrayList vp = new ArrayList<>();
// Inserting reverse with elements
// in the vector pair
for(int i = 0; i < n; i++)
{
vp.add(new int[]{reversDigits(arr[i]),
arr[i]});
}
// Sort the vector, this will sort the pair
// according to the reverse of elements
Collections.sort(vp, (a, b) -> a[0] - b[0]);
// Print the sorted vector content
for(int i = 0; i < vp.size(); i++)
System.out.print(vp.get(i)[1] + " ");
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 12, 10, 102, 31, 15 };
int n = arr.length;
sortArr(arr, n);
}
}
// This code is contributed by offbeat Python3
# Python3 implementation of the approach
# Function to return the
# reverse of n
def reversDigits(num) :
rev_num = 0;
while (num > 0) :
rev_num = rev_num * 10 + num % 10;
num = num // 10;
return rev_num;
# Function to sort the array according to
# the reverse of elements
def sortArr(arr, n) :
# Vector to store the reverse
# with respective elements
vp = [];
# Inserting reverse with elements
# in the vector pair
for i in range(n) :
vp.append((reversDigits(arr[i]),arr[i]));
# Sort the vector, this will sort the pair
# according to the reverse of elements
vp.sort()
# Print the sorted vector content
for i in range(len(vp)) :
print(vp[i][1],end= " ");
# Driver code
if __name__ == "__main__" :
arr = [ 12, 10, 102, 31, 15 ];
n = len(arr);
sortArr(arr, n);
# This code is contributed by AnkitRai01C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to return the
// reverse of n
static int reversDigits(int num)
{
int rev_num = 0;
while (num > 0)
{
rev_num = rev_num * 10 + num % 10;
num = num / 10;
}
return rev_num;
}
// Function to sort the array according to
// the reverse of elements
static void sortArr(int[] arr, int n)
{
// Vector to store the reverse
// with respective elements
List> vp = new List>();
// Inserting reverse with elements
// in the vector pair
for (int i = 0; i < n; i++)
{
vp.Add(new Tuple(reversDigits(arr[i]),arr[i]));
}
// Sort the vector, this will sort the pair
// according to the reverse of elements
vp.Sort();
// Print the sorted vector content
for (int i = 0; i < vp.Count; i++)
Console.Write(vp[i].Item2 + " ");
}
// Driver code
static void Main()
{
int[] arr = { 12, 10, 102, 31, 15 };
int n = arr.Length;
sortArr(arr, n);
}
}
// This code is contributed by divyesh072019 输出：
10 31 12 15 102时间复杂度：
其中N是数组的大小