编写一个函数来检查两个给定的字符串是否相同 相同字母异序词 不管是不是彼此。一个字符串的字谜是另一个包含相同字符的字符串，只有字符的顺序可以不同。例如，“abcd”和“dabc”是彼此的一个字谜。
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我们强烈建议您在继续解决方案之前单击此处并进行练习。
方法1（使用排序）
- 对两个字符串进行排序
- 比较排序后的字符串
以下是上述理念的实施情况：
C++
// C++ program to check whether two strings are anagrams // of each other #include <bits/stdc++.h> using namespace std; /* function to check whether two strings are anagram of each other */ bool areAnagram(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 anagram if (n1 != n2) return false ; // Sort both the 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 code int main() { string str1 = "test" ; string str2 = "ttew" ; // Function Call if (areAnagram(str1, str2)) cout << "The two strings are anagram of each other" ; else cout << "The two strings are not anagram of each " "other" ; return 0; } |
JAVA
// JAVA program to check whether two strings // are anagrams of each other import java.io.*; import java.util.Arrays; import java.util.Collections; class GFG { /* function to check whether two strings are anagram of each other */ static boolean areAnagram( char [] str1, char [] 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 anagram if (n1 != n2) return false ; // Sort both strings Arrays.sort(str1); Arrays.sort(str2); // Compare sorted strings for ( int i = 0 ; i < n1; i++) if (str1[i] != str2[i]) return false ; return true ; } /* Driver Code*/ public static void main(String args[]) { char str1[] = { 't' , 'e' , 's' , 't' }; char str2[] = { 't' , 't' , 'e' , 'w' }; // Function Call if (areAnagram(str1, str2)) System.out.println( "The two strings are" + " anagram of each other" ); else System.out.println( "The two strings are not" + " anagram of each other" ); } } // This code is contributed by Nikita Tiwari. |
python
# Python program to check whether two strings are # anagrams of each other # function to check whether two strings are anagram # of each other def areAnagram(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 anagram if n1 ! = n2: return 0 # Sort both strings str1 = sorted (str1) str2 = sorted (str2) # Compare sorted strings for i in range ( 0 , n1): if str1[i] ! = str2[i]: return 0 return 1 # Driver code str1 = "test" str2 = "ttew" # Function Call if areAnagram(str1, str2): print ( "The two strings are anagram of each other" ) else : print ( "The two strings are not anagram of each other" ) # This code is contributed by Bhavya Jain |
C#
// C# program to check whether two // strings are anagrams of each other using System; using System.Collections; class GFG { /* function to check whether two strings are anagram of each other */ public static bool areAnagram(ArrayList str1, ArrayList str2) { // Get lengths of both strings int n1 = str1.Count; int n2 = str2.Count; // If length of both strings is not // same, then they cannot be anagram if (n1 != n2) { return false ; } // Sort both strings str1.Sort(); str2.Sort(); // Compare sorted strings for ( int i = 0; i < n1; i++) { if (str1[i] != str2[i]) { return false ; } } return true ; } // Driver Code public static void Main( string [] args) { // create and initialize new ArrayList ArrayList str1 = new ArrayList(); str1.Add( 't' ); str1.Add( 'e' ); str1.Add( 's' ); str1.Add( 't' ); // create and initialize new ArrayList ArrayList str2 = new ArrayList(); str2.Add( 't' ); str2.Add( 't' ); str2.Add( 'e' ); str2.Add( 'w' ); // Function call if (areAnagram(str1, str2)) { Console.WriteLine( "The two strings are" + " anagram of each other" ); } else { Console.WriteLine( "The two strings are not" + " anagram of each other" ); } } } // This code is contributed by Shrikant13 |
Javascript
<script> // JavaScript program to check whether two strings // are anagrams of each other /* function to check whether two strings are anagram of each other */ function areAnagram(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 anagram if (n1 != n2) return false ; // Sort both strings str1.sort(); str2.sort() // Compare sorted strings for (let i = 0; i < n1; i++) if (str1[i] != str2[i]) return false ; return true ; } /* Driver Code*/ let str1=[ 't' , 'e' , 's' , 't' ]; let str2=[ 't' , 't' , 'e' , 'w' ]; // Function Call if (areAnagram(str1, str2)) document.write( "The two strings are" + " anagram of each other<br>" ); else document.write( "The two strings are not" + " anagram of each other<br>" ); // This code is contributed by rag2127 </script> |
输出：
The two strings are not anagram of each other
时间复杂性： O（nLogn）
方法2（计数字符） 此方法假定两个字符串中可能包含的字符集都很小。在下面的实现中，假设使用8位存储字符，可能有256个字符。
- 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。
- 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。
- 比较计数数组。如果两个计数数组相同，则返回true。
以下是上述理念的实施情况：
C++
// C++ program to check if two strings // are anagrams of each other #include <bits/stdc++.h> using namespace std; #define NO_OF_CHARS 256 /* function to check whether two strings are anagram of each other */ bool areAnagram( char * str1, char * 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 code*/ int main() { char str1[] = "geeksforgeeks" ; char str2[] = "forgeeksgeeks" ; // Function Call if (areAnagram(str1, str2)) cout << "The two strings are anagram of each other" ; else cout << "The two strings are not anagram of each " "other" ; return 0; } // This is code is contributed by rathbhupendra |
C
// C program to check if two strings // are anagrams of each other #include <stdio.h> #define NO_OF_CHARS 256 /* function to check whether two strings are anagram of each other */ bool areAnagram( char * str1, char * 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 code*/ int main() { char str1[] = "geeksforgeeks" ; char str2[] = "forgeeksgeeks" ; // Function Call if (areAnagram(str1, str2)) printf ( "The two strings are anagram of each other" ); else printf ( "The two strings are not anagram of each " "other" ); return 0; } |
JAVA
// JAVA program to check if two strings // are anagrams of each other import java.io.*; import java.util.*; class GFG { static int NO_OF_CHARS = 256 ; /* function to check whether two strings are anagram of each other */ static boolean areAnagram( 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 < 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*/ public static void main(String args[]) { char str1[] = ( "geeksforgeeks" ).toCharArray(); char str2[] = ( "forgeeksgeeks" ).toCharArray(); // Function call if (areAnagram(str1, str2)) System.out.println( "The two strings are" + "anagram of each other" ); else System.out.println( "The two strings are not" + " anagram of each other" ); } } // This code is contributed by Nikita Tiwari. |
python
# Python program to check if two strings are anagrams of # each other NO_OF_CHARS = 256 # Function to check whether two strings are anagram of # each other def areAnagram(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 code str1 = "geeksforgeeks" str2 = "forgeeksgeeks" # Function call if areAnagram(str1, str2): print "The two strings are anagram of each other" else : print "The two strings are not anagram of each other" # This code is contributed by Bhavya Jain |
C#
// C# program to check if two strings // are anagrams of each other using System; public class GFG { static int NO_OF_CHARS = 256; /* function to check whether two strings are anagram of each other */ static bool areAnagram( 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 < 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*/ public static void Main() { char [] str1 = ( "geeksforgeeks" ).ToCharArray(); char [] str2 = ( "forgeeksgeeks" ).ToCharArray(); // Function Call if (areAnagram(str1, str2)) Console.WriteLine( "The two strings are" + "anagram of each other" ); else Console.WriteLine( "The two strings are not" + " anagram of each other" ); } } // This code contributed by 29AjayKumar |
Javascript
<script> // JAVAscript program to check if two strings // are anagrams of each other let NO_OF_CHARS = 256; /* function to check whether two strings are anagram of each other */ function areAnagram(str1, str2) { // Create 2 count arrays and initialize // all values as 0 let count1 = new Array(NO_OF_CHARS); let count2 = new Array(NO_OF_CHARS); for (let i = 0; i < NO_OF_CHARS; i++) { count1[i] = 0; count2[i] = 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].charCodeAt(0)]++; count2[str1[i].charCodeAt(0)]++; } // 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( "" ); // Function call if (areAnagram(str1, str2)) document.write( "The two strings are" + "anagram of each other<br>" ); else document.write( "The two strings are not" + " anagram of each other<br>" ); // This code is contributed by avanitrachhadiya2155 </script> |
输出：
The two strings are anagram of each other
方法3（使用一个数组计数字符） 上述实现可以进一步简化为仅使用一个计数数组而不是两个。我们可以为str1中的字符增加计数数组中的值，为str2中的字符减少计数数组中的值。最后，如果所有的计数值都是0，那么这两个字符串就是彼此的字谜。幸亏 王牌 感谢您提出的优化建议。
C++
// C++ program to check if two strings // are anagrams of each other #include <bits/stdc++.h> using namespace std; #define NO_OF_CHARS 256 bool areAnagram( char * str1, char * 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 ; } // Driver code int main() { char str1[] = "geeksforgeeks" ; char str2[] = "forgeeksgeeks" ; // Function call if (areAnagram(str1, str2)) cout << "The two strings are anagram of each other" ; else cout << "The two strings are not anagram of each " "other" ; return 0; } |
JAVA
// Java program to check if two strings // are anagrams of each other class GFG{ static int NO_OF_CHARS = 256 ; // function to check if two strings // are anagrams of each other static boolean areAnagram( 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 ; i < str1.length; i++) { count[str1[i] - 'a' ]++; count[str2[i] - 'a' ]--; } // 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 ; // 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 ; } // Driver code public static void main(String[] args) { char str1[] = "geeksforgeeks" .toCharArray(); char str2[] = "forgeeksgeeks" .toCharArray(); // Function call if (areAnagram(str1, str2)) System.out.print( "The two strings are " + "anagram of each other" ); else System.out.print( "The two strings are " + "not anagram of each other" ); } } // This code is contributed by mark_85 |
Python3
# Python program to check if two strings # are anagrams of each other NO_OF_CHARS = 256 # function to check if two strings # are anagrams of each other def areAnagram(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 for i in range ( len (str1)): count[ ord (str1[i]) - ord ( 'a' )] + = 1 ; count[ ord (str2[i]) - ord ( 'a' )] - = 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 False ; # See if there is any non-zero # value in count array for i in range (NO_OF_CHARS): if (count[i] ! = 0 ): return False return True # Driver code str1 = "geeksforgeeks" str2 = "forgeeksgeeks" # Function call if (areAnagram(str1, str2)): print ( "The two strings are anagram of each other" ) else : print ( "The two strings are not anagram of each other" ) # This code is contributed by patel2127 |
C#
// C# program to check if two strings // are anagrams of each other using System; class GFG{ static int NO_OF_CHARS = 256; // function to check if two strings // are anagrams of each other static bool areAnagram( 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; i < str1.Length; i++) { count[str1[i] - 'a' ]++; count[str2[i] - 'a' ]--; } // 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 ; // 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 ; } // Driver code public static void Main(String []args) { char []str1 = "geeksforgeeks" .ToCharArray(); char []str2 = "forgeeksgeeks" .ToCharArray(); // Function call if (areAnagram(str1, str2)) Console.Write( "The two strings are " + "anagram of each other" ); else Console.Write( "The two strings are " + "not anagram of each other" ); } } // This code is contributed by shivanisinghss2110 |
Javascript
<script> // Javascript program to check if two strings // are anagrams of each other let NO_OF_CHARS = 256; // function to check if two strings // are anagrams of each other function areAnagram(str1, str2) { // Create a count array and initialize // all values as 0 let count = new Array(NO_OF_CHARS); for (let i = 0; i < NO_OF_CHARS; i++) { count[i] = 0; } let i; // For each character in input strings, // increment count in the corresponding // count array for (i = 0; i < str1.length; i++) { count[str1[i].charCodeAt(0) - 'a' .charCodeAt(0)]++; count[str2[i].charCodeAt(0) - 'a' .charCodeAt(0)]--; } // 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 ; // 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 ; } // Driver code let str1 = "geeksforgeeks" .split( "" ); let str2 = "forgeeksgeeks" .split( "" ); // Function call if (areAnagram(str1, str2)) document.write( "The two strings are " + "anagram of each other" ); else document.write( "The two strings are " + "not anagram of each other" ); // This code is contributed by unknown2108. </script> |
输出：
The two strings are anagram of each other
时间复杂性： O（n）
方法4（将所有字符放入HashMap）
在上面的实现中，我们在创建256个字符的数组时使用了额外的空间，但我们可以使用HashMap对其进行优化，我们可以在HashMap中存储字符和字符计数。其想法是将一个字符串的所有字符放入HashMap中，并在循环另一个字符串时减少它们。
JAVA
import java.io.*; import java.util.*; class GFG { public static boolean isAnagram(String a, String b) { // Check if length of both strings is same or not if (a.length() != b.length()) { return false ; } // Create a HashMap containing Character as Key and // Integer as Value. We will be storing character as // Key and count of character as Value. HashMap<Character, Integer> map = new HashMap<>(); // Loop over all character of String a and put in // HashMap. for ( int i = 0 ; i < a.length(); i++) { // Check if HashMap already contain current // character or not if (map.containsKey(a.charAt(i))) { // If contains increase count by 1 for that // character map.put(a.charAt(i), map.get(a.charAt(i)) + 1 ); } else { // else put that character in map and set // count to 1 as character is encountered // first time map.put(a.charAt(i), 1 ); } } // Now loop over String b for ( int i = 0 ; i < b.length(); i++) { // Check if current character already exists in // HashMap/map if (map.containsKey(b.charAt(i))) { // If contains reduce count of that // character by 1 to indicate that current // character has been already counted as // idea here is to check if in last count of // all characters in last is zero which // means all characters in String a are // present in String b. map.put(b.charAt(i), map.get(b.charAt(i)) - 1 ); } } // Extract all keys of HashMap/map Set<Character> keys = map.keySet(); // Loop over all keys and check if all keys are 0. // If so it means it is anagram. for (Character key : keys) { if (map.get(key) != 0 ) { return false ; } } // Returning True as all keys are zero return true ; } public static void main(String[] args) { String str1 = "geeksforgeeks" ; String str2 = "forgeeksgeeks" ; // Function call if (isAnagram(str1, str2)) System.out.print( "The two strings are " + "anagram of each other" ); else System.out.print( "The two strings are " + "not anagram of each other" ); } } |
C#
using System; using System.Collections.Generic; public class GFG { public static bool isAnagram(String a, String b) { // Check if length of both strings is same or not if (a.Length != b.Length) { return false ; } // Create a Dictionary containing char as Key and // int as Value. We will be storing character as // Key and count of character as Value. Dictionary< char , int > map = new Dictionary< char , int >(); // Loop over all character of String a and put in // Dictionary. for ( int i = 0; i < a.Length; i++) { // Check if Dictionary already contain current // character or not if (map.ContainsKey(a[i])) { // If contains increase count by 1 for that // character map[a[i]] = map[a[i]] + 1; } else { // else put that character in map and set // count to 1 as character is encountered // first time map.Add(a[i], 1); } } // Now loop over String b for ( int i = 0; i < b.Length; i++) { // Check if current character already exists in // Dictionary/map if (map.ContainsKey(b[i])) { // If contains reduce count of that // character by 1 to indicate that current // character has been already counted as // idea here is to check if in last count of // all characters in last is zero which // means all characters in String a are // present in String b. map[b[i]]= map[b[i]] - 1; } } // Extract all keys of Dictionary/map var keys = map.Keys; // Loop over all keys and check if all keys are 0. // If so it means it is anagram. foreach ( char key in keys) { if (map[key] != 0) { return false ; } } // Returning True as all keys are zero return true ; } // Driver code public static void Main(String[] args) { String str1 = "geeksforgeeks" ; String str2 = "forgeeksgeeks" ; // Function call if (isAnagram(str1, str2)) Console.Write( "The two strings are " + "anagram of each other" ); else Console.Write( "The two strings are " + "not anagram of each other" ); } } // This code is contributed by shikhasingrajput |
输出：
The two strings are anagram of each other
时间复杂性： O（n）
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