具有唯一（或不同）数位的数字
原文：https://www.geeksforgeeks.org/numbers-unique-distinct-digits/
给定范围，请打印所有具有唯一数字的数字。
示例：
Input : 10 20
Output : 10 12 13 14 15 16 17 18 19 20 (Except 11)
Input : 1 10
Output : 1 2 3 4 5 6 7 8 9 10
As the problem is pretty simple, the only thing to be done is :-
1- Find the digits one by one and keep marking visited digits.
2- If all digits occurs one time only then print that number.
3- Else not.
C++
// C++ implementation to find unique digit
// numbers in a range
#include<bits/stdc++.h>
using namespace std;
// Function to print unique digit numbers
// in range from l to r.
void printUnique(int l, int r)
{
// Start traversing the numbers
for (int i=l ; i<=r ; i++)
{
int num = i;
bool visited[10] = {false};
// Find digits and maintain its hash
while (num)
{
// if a digit occurs more than 1 time
// then break
if (visited[num % 10])
break;
visited[num%10] = true;
num = num/10;
}
// num will be 0 only when above loop
// doesn't get break that means the
// number is unique so print it.
if (num == 0)
cout << i << " ";
}
}
// Driver code
int main()
{
int l = 1, r = 20;
printUnique(l, r);
return 0;
}
Java
// Java implementation to find unique digit
// numbers in a range
class Test
{
// Method to print unique digit numbers
// in range from l to r.
static void printUnique(int l, int r)
{
// Start traversing the numbers
for (int i=l ; i<=r ; i++)
{
int num = i;
boolean visited[] = new boolean[10];
// Find digits and maintain its hash
while (num != 0)
{
// if a digit occurs more than 1 time
// then break
if (visited[num % 10])
break;
visited[num%10] = true;
num = num/10;
}
// num will be 0 only when above loop
// doesn't get break that means the
// number is unique so print it.
if (num == 0)
System.out.print(i + " ");
}
}
// Driver method
public static void main(String args[])
{
int l = 1, r = 20;
printUnique(l, r);
}
}
Python3
# Python3 implementation
# to find unique digit
# numbers in a range
# Function to print
# unique digit numbers
# in range from l to r.
def printUnique(l,r):
# Start traversing
# the numbers
for i in range (l, r + 1):
num = i;
visited = [0,0,0,0,0,0,0,0,0,0];
# Find digits and
# maintain its hash
while (num):
# if a digit occurs
# more than 1 time
# then break
if visited[num % 10] == 1:
break;
visited[num % 10] = 1;
num = (int)(num / 10);
# num will be 0 only when
# above loop doesn't get
# break that means the
# number is unique so
# print it.
if num == 0:
print(i, end = " ");
# Driver code
l = 1;
r = 20;
printUnique(l, r);
# This code is
# contributed by mits
C
// C# implementation to find unique digit
// numbers in a range
using System;
public class GFG {
// Method to print unique digit numbers
// in range from l to r.
static void printUnique(int l, int r)
{
// Start traversing the numbers
for (int i = l ; i <= r ; i++)
{
int num = i;
bool []visited = new bool[10];
// Find digits and maintain
// its hash
while (num != 0)
{
// if a digit occurs more
// than 1 time then break
if (visited[num % 10])
break;
visited[num % 10] = true;
num = num / 10;
}
// num will be 0 only when
// above loop doesn't get
// break that means the number
// is unique so print it.
if (num == 0)
Console.Write(i + " ");
}
}
// Driver method
public static void Main()
{
int l = 1, r = 20;
printUnique(l, r);
}
}
// This code is contributed by Sam007\.
PHP
<?php
// PHP implementation to find unique
// digit numbers in a range
// Function to print unique digit
// numbers in range from l to r.
function printUnique($l, $r)
{
// Start traversing the numbers
for ($i = $l ; $i <= $r; $i++)
{
$num = $i;
$visited = (false);
// Find digits and
// maintain its hash
while ($num)
{
// if a digit occurs more
// than 1 time then break
if ($visited[$num % 10])
$visited[$num % 10] = true;
$num = (int)$num / 10;
}
// num will be 0 only when above
// loop doesn't get break that
// means the number is unique
// so print it.
if ($num == 0)
echo $i , " ";
}
}
// Driver code
$l = 1; $r = 20;
printUnique($l, $r);
// This code is contributed by aj_36
?>
输出：
1 2 3 4 5 6 7 8 9 10 12 13 14 15 16 17 18 19 20
本文由 Sahil Chhabra 贡献。 如果您喜欢 GeeksforGeeks 并希望做出贡献，则还可以使用 tribution.geeksforgeeks.org 撰写文章，或将您的文章邮寄至 tribution@geeksforgeeks.org。 查看您的文章出现在 GeeksforGeeks 主页上，并帮助其他 Geeks。
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