给定两个整数A和B ，任务是检查给定的数字是否是彼此的字谜。就像字符串一样，如果一个数字可以通过改变其中的数字使其与另一个数字相等，则称该数字是某个其他数字的变位词。
例子：
Input: A = 204, B = 240
Output: Yes
Input: A = 23, B = 959
Output: No
方法：创建两个数组freqA[]和freqB[] ，其中freqA[i]和freqB[i]将分别存储数字i在a和b 中的频率。现在遍历频率数组，对于任何数字i，如果freqA[i] != freqB[i]则这些数字不是彼此的字谜。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
const int TEN = 10;
// Function to update the frequency array
// such that freq[i] stores the
// frequency of digit i in n
void updateFreq(int n, int freq[])
{
// While there are digits
// left to process
while (n) {
int digit = n % TEN;
// Update the frequency of
// the current digit
freq[digit]++;
// Remove the last digit
n /= TEN;
}
}
// Function that returns true if a and b
// are anagarams of each other
bool areAnagrams(int a, int b)
{
// To store the frequencies of
// the digits in a and b
int freqA[TEN] = { 0 };
int freqB[TEN] = { 0 };
// Update the frequency of
// the digits in a
updateFreq(a, freqA);
// Update the frequency of
// the digits in b
updateFreq(b, freqB);
// Match the frequencies of
// the common digits
for (int i = 0; i < TEN; i++) {
// If frequency differs for any digit
// then the numbers are not
// anagrams of each other
if (freqA[i] != freqB[i])
return false;
}
return true;
}
// Driver code
int main()
{
int a = 240, b = 204;
if (areAnagrams(a, b))
cout << "Yes";
else
cout << "No";
return 0;
} Java
// Java implementation of the approach
class GFG
{
static final int TEN = 10;
// Function to update the frequency array
// such that freq[i] stores the
// frequency of digit i in n
static void updateFreq(int n, int [] freq)
{
// While there are digits
// left to process
while (n > 0)
{
int digit = n % TEN;
// Update the frequency of
// the current digit
freq[digit]++;
// Remove the last digit
n /= TEN;
}
}
// Function that returns true if a and b
// are anagarams of each other
static boolean areAnagrams(int a, int b)
{
// To store the frequencies of
// the digits in a and b
int [] freqA = new int[TEN];
int [] freqB = new int[TEN];
// Update the frequency of
// the digits in a
updateFreq(a, freqA);
// Update the frequency of
// the digits in b
updateFreq(b, freqB);
// Match the frequencies of
// the common digits
for (int i = 0; i < TEN; i++)
{
// If frequency differs for any digit
// then the numbers are not
// anagrams of each other
if (freqA[i] != freqB[i])
return false;
}
return true;
}
// Driver code
public static void main (String[] args)
{
int a = 204, b = 240;
if (areAnagrams(a, b))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by ihirtikPython3
# Python3 implementation of the approach
TEN = 10
# Function to update the frequency array
# such that freq[i] stores the
# frequency of digit i in n
def updateFreq(n, freq) :
# While there are digits
# left to process
while (n) :
digit = n % TEN
# Update the frequency of
# the current digit
freq[digit] += 1
# Remove the last digit
n //= TEN
# Function that returns true if a and b
# are anagarams of each other
def areAnagrams(a, b):
# To store the frequencies of
# the digits in a and b
freqA = [ 0 ] * TEN
freqB = [ 0 ] * TEN
# Update the frequency of
# the digits in a
updateFreq(a, freqA)
# Update the frequency of
# the digits in b
updateFreq(b, freqB)
# Match the frequencies of
# the common digits
for i in range(TEN):
# If frequency differs for any digit
# then the numbers are not
# anagrams of each other
if (freqA[i] != freqB[i]):
return False
return True
# Driver code
a = 240
b = 204
if (areAnagrams(a, b)):
print("Yes")
else:
print("No")
# This code is contributed by
# divyamohan123C#
// C# implementation of the approach
using System;
class GFG
{
static int TEN = 10;
// Function to update the frequency array
// such that freq[i] stores the
// frequency of digit i in n
static void updateFreq(int n, int [] freq)
{
// While there are digits
// left to process
while (n > 0)
{
int digit = n % TEN;
// Update the frequency of
// the current digit
freq[digit]++;
// Remove the last digit
n /= TEN;
}
}
// Function that returns true if a and b
// are anagarams of each other
static bool areAnagrams(int a, int b)
{
// To store the frequencies of
// the digits in a and b
int [] freqA = new int[TEN];
int [] freqB = new int[TEN];;
// Update the frequency of
// the digits in a
updateFreq(a, freqA);
// Update the frequency of
// the digits in b
updateFreq(b, freqB);
// Match the frequencies of
// the common digits
for (int i = 0; i < TEN; i++)
{
// If frequency differs for any digit
// then the numbers are not
// anagrams of each other
if (freqA[i] != freqB[i])
return false;
}
return true;
}
// Driver code
public static void Main ()
{
int a = 204, b = 240;
if (areAnagrams(a, b))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by ihirtikJavascript
Python3
# Python3 implementation of the approach
# Function that returns true if a and b
# are anagarams of each other
def areAnagrams(a, b):
# Converting numbers to strings
a = str(a)
b = str(b)
# Checking if the sorting values
# of two strings are equal
if(sorted(a) == sorted(b)):
return True
else:
return False
# Driver code
a = 240
b = 204
if (areAnagrams(a, b)):
print("Yes")
else:
print("No")
# This code is contributed by vikkycirus输出：
Yes方法#2：使用排序（）：
- 将两个整数转换为字符串。
- 对字符串排序并检查它们是否相等。
- 如果它们相等，则打印 Yes。
- 否则没有。
下面是实现：
蟒蛇3
# Python3 implementation of the approach
# Function that returns true if a and b
# are anagarams of each other
def areAnagrams(a, b):
# Converting numbers to strings
a = str(a)
b = str(b)
# Checking if the sorting values
# of two strings are equal
if(sorted(a) == sorted(b)):
return True
else:
return False
# Driver code
a = 240
b = 204
if (areAnagrams(a, b)):
print("Yes")
else:
print("No")
# This code is contributed by vikkycirus
输出：
Yes如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。