按排序顺序打印给定数字中的所有重复数字
原文:https://www . geesforgeks . org/print-all-repeating-digits-present-in-给定的数字-按排序顺序/
给定一个整数 N ，任务是按排序顺序打印所有出现在 N 中的重复数字。
示例:
输入: N = 45244336543 输出: 3 4 5 说明:重复数字为 3 4 5
输入: N = 987065467809 输出: 0 6 7 8 9
方式:思路是用哈希存储 N 的数字的频率，打印那些频率超过 1 的数字。
- 初始化一个 HashMap ，存储数字 0-9 的频率。
- 将整数转换为其等价字符串。
- 迭代字符串的字符，对于每个字符，执行以下步骤:
- 遍历 HashMap 打印计数超过 1 的数字。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to print repeating
// digits present in the number N
void printDup(string N)
{
// Stores the count of
// unique digits
int res = 0;
// Map to store frequency
// of each digit
int cnt[10];
// Set count of all digits to 0
for (int i = 0; i < 10; i++)
cnt[i] = 0;
// Traversing the string
for (int i = 0; i < N.size(); i++) {
// Convert the character
// to equivalent integer
int digit = (N[i] - '0');
// Increase the count of digit
cnt[digit] += 1;
}
// Traverse the Map
for (int i = 0; i < 10; i++) {
// If frequency
// of digit exceeds 1
if (cnt[i] > 1)
cout << i << " ";
}
cout << endl;
}
// Driver Code
int main()
{
string N = "45244336543";
// Function call to print
// repeating digits in N
printDup(N);
return 0;
}
// This code is contributed by Kingash.
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG {
// Function to print repeating
// digits present in the number N
static void printDup(String N)
{
// Stores the count of
// unique digits
int res = 0;
// Map to store frequency
// of each digit
int cnt[] = new int[10];
// Traversing the string
for (int i = 0; i < N.length(); i++) {
// Convert the character
// to equivalent integer
int digit = (N.charAt(i) - '0');
// Increase the count of digit
cnt[digit] += 1;
}
// Traverse the Map
for (int i = 0; i < 10; i++) {
// If frequency
// of digit exceeds 1
if (cnt[i] > 1)
System.out.print(i + " ");
}
System.out.println();
}
// Driver Code
public static void main(String[] args)
{
String N = "45244336543";
// Function call to print
// repeating digits in N
printDup(N);
}
}
// This code is contributed by Kingash.
Python 3
# Python implementation
# of the above approach
# Function to print repeating
# digits present in the number N
def printDup(N):
# Stores the count of
# unique digits
res = 0
# Set count of all digits to 0
cnt = [0] * 10
# Convert integer to
# equivalent string
string = str(N)
# Traversing the string
for i in string:
# Convert the character
# to equivalent integer
digit = int(i)
# Increase the count of digit
cnt[digit] += 1
# Traverse the Map
for i in range(10):
# If frequency
# of digit is 1
if (cnt[i] > 1):
# If count exceeds 1
print(i, end=" ")
# Driver Code
N = 45244336543
# Function call to print
# repeating digits in N
printDup(N)
C
// C# program to implement
// the above approach
using System;
class GFG
{
// Function to print repeating
// digits present in the number N
static void printDup(string N)
{
// Stores the count of
// unique digits
int res = 0;
// Map to store frequency
// of each digit
int[] cnt = new int[10];
// Traversing the string
for (int i = 0; i < N.Length; i++) {
// Convert the character
// to equivalent integer
int digit = (N[i] - '0');
// Increase the count of digit
cnt[digit] += 1;
}
// Traverse the Map
for (int i = 0; i < 10; i++) {
// If frequency
// of digit exceeds 1
if (cnt[i] > 1)
Console.Write(i + " ");
}
Console.WriteLine();
}
// Driver code
public static void Main()
{
string N = "45244336543";
// Function call to print
// repeating digits in N
printDup(N);
}
}
// This code is contributed by code_hunt.
java 描述语言
<script>
// Javascript program for the above approach
// Function to print repeating
// digits present in the number N
function printDup(N)
{
// Stores the count of
// unique digits
var res = 0;
// Map to store frequency
// of each digit
var cnt = Array(10);
// Set count of all digits to 0
for (var i = 0; i < 10; i++)
cnt[i] = 0;
// Traversing the string
for (var i = 0; i < N.length; i++) {
// Convert the character
// to equivalent integer
var digit = (N[i] - '0');
// Increase the count of digit
cnt[digit] += 1;
}
// Traverse the Map
for (var i = 0; i < 10; i++) {
// If frequency
// of digit exceeds 1
if (cnt[i] > 1)
document.write( i + " ");
}
document.write("<br>");
}
// Driver Code
var N = "45244336543";
// Function call to print
// repeating digits in N
printDup(N);
</script>
Output:
3 4 5
时间复杂度:O(log10N) 辅助空间: O(1)
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