本文概述

• C ++
• Java
• python
• C#
• C ++
• C
• Java
• python
• C#
• CPP
• C ++

编写函数以检查两个给定的字符串是否为字谜彼此之间。字符串的字谜是另一个包含相同字符的字符串, 只有字符顺序可以不同。例如, " abcd"和" dabc"是彼此的字谜。

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【算法题（检查两个字符串是否互为字谜）】方法1(使用排序)

1. 对两个字符串进行排序
2. 比较排序的字符串

以下是上述想法的实现：
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 lenghts 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 lenght 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 lenghts 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 initalize new ArrayList ArrayList str1 = new ArrayList(); str1.Add( 't' ); str1.Add( 'e' ); str1.Add( 's' ); str1.Add( 't' ); //create and initalize 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

输出如下：

The two strings are not anagram of each other

时间复杂度：O(nLogn)
方法2(计数字符)
此方法假定两个字符串中的可能字符集很小。在以下实现中, 假定使用8位存储字符, 并且可以有256个可能的字符。

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, 并递增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。

以下是上述想法的实现：
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[] = "lsbin" ; 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[] = "lsbin" ; 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[] = ( "lsbin" ).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 = "lsbin" 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 = ( "lsbin" ).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

输出如下：

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是仅使用一个计数数组而不是两个。我们可以在count数组中为str1中的字符增加值, 并为str2中的字符减少。最后, 如果所有计数值均为0, 则两个字符串彼此相似。谢谢
高手
建议这种优化。
CPP

//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[] = "lsbin" ; 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; }

时间复杂度：O(n)
方法4(求和)
该问题可以在线性时间和恒定空间中完成。

1. 我们将变量数count初始化为0。
2. 然后, 我们取第一个字符串的所有字符的总和, 然后从第二个字符串减小所有字符的值。
3. 如果Count值最终为0, 即在执行任何操作之前, 则为anagram, 否则为0。

下面是上述方法的实现：
C ++

//C++ program to check if two strings //are anagrams of each other #include < bits/stdc++.h> using namespace std; bool isAnagram(string c, string d) { if (c.size() != d.size()) return false ; int count = 0; //Take sum of all characters of first String for ( int i = 0; i < c.size(); i++) { count += c[i]; } //Subtract the Value of all the characters of second //String for ( int i = 0; i < d.size(); i++) { count -= d[i]; } //If Count = 0 then they are anagram //If count> 0 or count < 0 then they are not anagram return (count == 0); } //Driver code int main() { char str1[] = "lsbin" ; char str2[] = "forgeeksgeeks" ; //Function call if (isAnagram(str1, str2)) cout < < "The two strings are anagram of each other" ; else cout < < "The two strings are not anagram of each " "other" ; return 0; }

输出如下

The two strings are anagram of each other

时间复杂度：O(n)
辅助空间：O(1)
如果发现任何不正确的地方, 或者想分享有关上述主题的更多信息, 请写评论。

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