检查两个字符串是否相互排列

编写一个函数来检查两个给定的字符串是否相互排列。字符串的排列是另一个包含相同字符的字符串，只是字符的顺序可以不同。例如，“abcd”和“dabc”是彼此的排列。

我们强烈建议您单击此处并进行练习，然后再继续使用解决方案。

方法一（使用排序）
1）对两个字符串进行排序
2) 比较排序后的字符串

C++

// C++ program to check whether two strings are
// Permutations of each other
#include 
using namespace std;
 
/* function to check whether two strings are
   Permutation of each other */
bool arePermutation(string str1, string str2)
{
    // Get lengths of both strings
    int n1 = str1.length();
    int n2 = str2.length();
 
    // If length of both strings is not same,
    // then they cannot be Permutation
    if (n1 != n2)
      return false;
 
    // Sort both strings
    sort(str1.begin(), str1.end());
    sort(str2.begin(), str2.end());
 
    // Compare sorted strings
    for (int i = 0; i < n1;  i++)
       if (str1[i] != str2[i])
         return false;
 
    return true;
}
 
/* Driver program to test to print printDups*/
int main()
{
    string str1 = "test";
    string str2 = "ttew";
    if (arePermutation(str1, str2))
      printf("Yes");
    else
      printf("No");
 
    return 0;
}

Java

// Java program to check whether two strings are
// Permutations of each other
import java.util.*;
class GfG {
 
/* function to check whether two strings are
Permutation of each other */
static boolean arePermutation(String str1, String str2)
{
    // Get lengths of both strings
    int n1 = str1.length();
    int n2 = str2.length();
 
    // If length of both strings is not same,
    // then they cannot be Permutation
    if (n1 != n2)
    return false;
    char ch1[] = str1.toCharArray();
    char ch2[] = str2.toCharArray();
 
    // Sort both strings
    Arrays.sort(ch1);
    Arrays.sort(ch2);
 
    // Compare sorted strings
    for (int i = 0; i < n1; i++)
    if (ch1[i] != ch2[i])
        return false;
 
    return true;
}
 
/* Driver program to test to print printDups*/
public static void main(String[] args)
{
    String str1 = "test";
    String str2 = "ttew";
    if (arePermutation(str1, str2))
    System.out.println("Yes");
    else
    System.out.println("No");
 
}
}

Python3

# Python3 program to check whether two
# strings are Permutations of each other
 
# function to check whether two strings
# are Permutation of each other */
def arePermutation(str1, str2):
     
    # Get lengths of both strings
    n1 = len(str1)
    n2 = len(str2)
 
    # If length of both strings is not same,
    # then they cannot be Permutation
    if (n1 != n2):
        return False
 
    # Sort both strings
    a = sorted(str1)
    str1 = " ".join(a)
    b = sorted(str2)
    str2 = " ".join(b)
 
    # Compare sorted strings
    for i in range(0, n1, 1):
        if (str1[i] != str2[i]):
            return False
 
    return True
 
# Driver Code
if __name__ == '__main__':
    str1 = "test"
    str2 = "ttew"
    if (arePermutation(str1, str2)):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by
# Sahil_Shelangia

C#

// C# program to check whether two strings are
// Permutations of each other
using System;
 
class GfG
{
 
/* function to check whether two strings are
Permutation of each other */
static bool arePermutation(String str1, String str2)
{
    // Get lengths of both strings
    int n1 = str1.Length;
    int n2 = str2.Length;
 
    // If length of both strings is not same,
    // then they cannot be Permutation
    if (n1 != n2)
        return false;
    char []ch1 = str1.ToCharArray();
    char []ch2 = str2.ToCharArray();
 
    // Sort both strings
    Array.Sort(ch1);
    Array.Sort(ch2);
 
    // Compare sorted strings
    for (int i = 0; i < n1; i++)
        if (ch1[i] != ch2[i])
            return false;
 
    return true;
}
 
/* Driver code*/
public static void Main(String[] args)
{
    String str1 = "test";
    String str2 = "ttew";
    if (arePermutation(str1, str2))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
 
// This code contributed by Rajput-Ji

Javascript

 
// Javascript program to check whether two
// strings are Permutations of each other
 
// Function to check whether two strings are
// Permutation of each other
function arePermutation(str1, str2)
{
     
    // Get lengths of both strings
    let n1 = str1.length;
    let n2 = str2.length;
 
    // If length of both strings is not same,
    // then they cannot be Permutation
    if (n1 != n2)
        return false;
         
    let ch1 = str1.split(' ');
    let ch2 = str2.split(' ');
 
    // Sort both strings
    ch1.sort();
    ch2.sort();
 
    // Compare sorted strings
    for(let i = 0; i < n1; i++)
        if (ch1[i] != ch2[i])
            return false;
 
    return true;
}
 
// Driver Code
let str1 = "test";
let str2 = "ttew";
 
if (arePermutation(str1, str2))
    document.write("Yes");
else
    document.write("No");
 
// This code is contributed by suresh07
 

---

---

输出：

No

时间复杂度：此方法的时间复杂度取决于所使用的排序技术。在上述实现中，使用了快速排序，在最坏的情况下可能为 O(n^2)。如果我们使用像归并排序这样的 O(nLogn) 排序算法，那么复杂度变为 O(nLogn)

方法2（计数字符）
此方法假定两个字符串中可能的字符集都很小。在下面的实现中，假设字符使用 8 位存储，可能有 256 个字符。
1) 为两个字符串创建大小为 256 的计数数组。将计数数组中的所有值初始化为 0。
2）遍历两个字符串的每个字符并增加相应计数数组中的字符计数。
3) 比较计数数组。如果两个计数数组相同，则返回 true。

C++

// C++ program to check whether two strings are
// Permutations of each other
#include 
using namespace std;
# define NO_OF_CHARS 256
 
/* function to check whether two strings are
   Permutation of each other */
bool arePermutation(string str1, string str2)
{
    // Create 2 count arrays and initialize
    // all values as 0
    int count1[NO_OF_CHARS] = {0};
    int count2[NO_OF_CHARS] = {0};
    int i;
 
    // For each character in input strings,
    // increment count in the corresponding
    // count array
    for (i = 0; str1[i] && str2[i];  i++)
    {
        count1[str1[i]]++;
        count2[str2[i]]++;
    }
 
    // If both strings are of different length.
    // Removing this condition will make the
    // program fail for strings like "aaca"
    // and "aca"
    if (str1[i] || str2[i])
      return false;
 
    // Compare count arrays
    for (i = 0; i < NO_OF_CHARS; i++)
        if (count1[i] != count2[i])
            return false;
 
    return true;
}
 
/* Driver program to test to print printDups*/
int main()
{
    string str1 = "geeksforgeeks";
    string str2 = "forgeeksgeeks";
    if ( arePermutation(str1, str2) )
      printf("Yes");
    else
      printf("No");
 
    return 0;
}

Java

// JAVA program to check if two strings
// are Permutations of each other
import java.io.*;
import java.util.*;
 
class GFG{
     
    static int NO_OF_CHARS = 256;
     
    /* function to check whether two strings
    are Permutation of each other */
    static boolean arePermutation(char str1[], char str2[])
    {
        // Create 2 count arrays and initialize
        // all values as 0
        int count1[] = new int [NO_OF_CHARS];
        Arrays.fill(count1, 0);
        int count2[] = new int [NO_OF_CHARS];
        Arrays.fill(count2, 0);
        int i;
  
        // For each character in input strings,
        // increment count in the corresponding
        // count array
        for (i = 0; i 

Python

# Python program to check if two strings are
# Permutations of each other
NO_OF_CHARS = 256
 
# Function to check whether two strings are
# Permutation of each other
def arePermutation(str1, str2):
 
    # Create two count arrays and initialize
    # all values as 0
    count1 = [0] * NO_OF_CHARS
    count2 = [0] * NO_OF_CHARS
 
    # For each character in input strings,
    # increment count in the corresponding
    # count array
    for i in str1:
        count1[ord(i)]+=1
 
    for i in str2:
        count2[ord(i)]+=1
 
    # If both strings are of different length.
    # Removing this condition will make the
    # program fail for strings like
    # "aaca" and "aca"
    if len(str1) != len(str2):
        return 0
 
    # Compare count arrays
    for i in xrange(NO_OF_CHARS):
        if count1[i] != count2[i]:
            return 0
 
    return 1
 
# Driver program to test the above functions
str1 = "geeksforgeeks"
str2 = "forgeeksgeeks"
if arePermutation(str1, str2):
    print "Yes"
else:
    print "No"
 
# This code is contributed by Bhavya Jain

C＃

// C# program to check if two strings
// are Permutations of each other
using System;
class GFG{
     
    static int NO_OF_CHARS = 256;
     
    /* function to check whether two strings
    are Permutation of each other */
    static bool arePermutation(char []str1, char []str2)
    {
        // Create 2 count arrays and initialize
        // all values as 0
        int []count1 = new int [NO_OF_CHARS];
        int []count2 = new int [NO_OF_CHARS];
        int i;
 
        // For each character in input strings,
        // increment count in the corresponding
        // count array
        for (i = 0; i 

Javascript

 
// Javascript program to check if two strings
// are Permutations of each other
let NO_OF_CHARS = 256;
  
/* Function to check whether two strings
are Permutation of each other */
function arePermutation(str1, str2)
{
     
    // Create 2 count arrays and initialize
    // all values as 0
    let count1 = Array(NO_OF_CHARS);
    let count2 = Array(NO_OF_CHARS);
    count1.fill(0);
    count2.fill(0);
    let i;
 
    // For each character in input strings,
    // increment count in the corresponding
    // count array
    for(i = 0;
        i < str1.length && i < str2.length;
        i++)
    {
        count1[str1[i]]++;
        count2[str2[i]]++;
    }
 
    // If both strings are of different length.
    // Removing this condition will make the program
    // fail for strings like "aaca" and "aca"
    if (str1.length != str2.length)
        return false;
 
    // Compare count arrays
    for(i = 0; i < NO_OF_CHARS; i++)
        if (count1[i] != count2[i])
            return false;
 
    return true;
}
 
// Driver code  
let str1 = ("geeksforgeeks").split('');
let str2 = ("forgeeksgeeks").split('');
 
if (arePermutation(str1, str2) )
    document.write("Yes");
else
    document.write("No");
     
// This code is contributed by rameshtravel07
 

输出：

Yes

上述实现可以进一步只使用一个计数数组而不是两个。我们可以为 str1 中的字符递增 count 数组中的值，为 str2 中的字符递减。最后，如果所有的计数值都为0，那么这两个字符串是彼此的Permutation。感谢 Ace 提出此优化建议。

C++

// C++ function to check whether two strings are
// Permutations of each other
bool arePermutation(string str1, string str2)
{
    // Create a count array and initialize all
    // values as 0
    int count[NO_OF_CHARS] = {0};
    int i;
 
    // For each character in input strings,
    // increment count in the corresponding
    // count array
    for (i = 0; str1[i] && str2[i];  i++)
    {
        count[str1[i]]++;
        count[str2[i]]--;
    }
 
    // If both strings are of different length.
    // Removing this condition  will make the
    // program fail for strings like "aaca" and
    // "aca"
    if (str1[i] || str2[i])
      return false;
 
    // See if there is any non-zero value in
    // count array
    for (i = 0; i < NO_OF_CHARS; i++)
        if (count[i])
            return false;
     return true;
}

Java

// Java function to check whether two strings are 
// Permutations of each other
static boolean arePermutation(char str1[], char str2[])
{
    
    // Create a count array and initialize all
    // values as 0
    int count[] = new int[NO_OF_CHARS];
    int i;
    
    // For each character in input strings, 
    // increment count in the corresponding 
    // count array
    for (i = 0; str1[i] && str2[i];  i++)
    {
        count[str1[i]]++;
        count[str2[i]]--;
    }
    
    // If both strings are of different length.
    // Removing this condition  will make the
    // program fail for strings like "aaca" and
    // "aca"
    if (str1[i] || str2[i])
      return false;
    
    // See if there is any non-zero value in 
    // count array
    for (i = 0; i < NO_OF_CHARS; i++)
        if (count[i] != 0)
            return false;
     return true;
}
 
// This code is contributed by divyesh072019.

Python3

# Python3 function to check whether two strings are
# Permutations of each other
def arePermutation(str1, str2):
 
    # Create a count array and initialize all
    # values as 0
    count = [0 for i in range(NO_OF_CHARS)]   
    i = 0
 
    # For each character in input strings,
    # increment count in the corresponding
    # count array
    while(str1[i] and str2[i]):
     
        count[str1[i]] += 1
        count[str2[i]] -= 1
 
    # If both strings are of different length.
    # Removing this condition  will make the
    # program fail for strings like "aaca" and
    # "aca"
    if (str1[i] or str2[i]):
      return False;
 
    # See if there is any non-zero value in
    # count array
    for i in range(NO_OF_CHARS):
        if (count[i]):
            return False;
    return True;
 
# This code is contributed by pratham76.

C＃

// C# function to check whether two strings are 
// Permutations of each other
static bool arePermutation(char[] str1, char[] str2)
{
   
    // Create a count array and initialize all
    // values as 0
    int[] count = new int[NO_OF_CHARS];
    int i;
   
    // For each character in input strings, 
    // increment count in the corresponding 
    // count array
    for (i = 0; str1[i] && str2[i];  i++)
    {
        count[str1[i]]++;
        count[str2[i]]--;
    }
   
    // If both strings are of different length.
    // Removing this condition  will make the
    // program fail for strings like "aaca" and
    // "aca"
    if (str1[i] || str2[i])
      return false;
   
    // See if there is any non-zero value in 
    // count array
    for (i = 0; i < NO_OF_CHARS; i++)
        if (count[i] != 0)
            return false;
     return true;
}
 
// This code is contributed by divyeshrabadiya07.

Javascript

 
// Javascript function to check whether two strings are
// Permutations of each other
 
function arePermutation(str1,str2)
{
    // Create a count array and initialize all
    // values as 0
    let count = new Array(NO_OF_CHARS);
    let i;
     
    // For each character in input strings,
    // increment count in the corresponding
    // count array
    for (i = 0; str1[i] && str2[i];  i++)
    {
        count[str1[i]]++;
        count[str2[i]]--;
    }
     
    // If both strings are of different length.
    // Removing this condition  will make the
    // program fail for strings like "aaca" and
    // "aca"
    if (str1[i] || str2[i])
      return false;
     
    // See if there is any non-zero value in
    // count array
    for (i = 0; i < NO_OF_CHARS; i++)
        if (count[i] != 0)
            return false;
     return true;
}
 
// This code is contributed by patel2127

如果可能的字符集只包含英文字母，那么我们可以将数组的大小减少到 58，并使用str[i] – 'A'作为计数数组的索引，因为 'A' 的 ASCII 值是 65 ， 'B'是 66, ..... , Z 是 90 并且 'a' 是 97 , 'b' 是 98 , ...... , 'z' 是 122。这将进一步优化此方法。