C++ 程序检查两个字符串是否互为 Anagram

编写一个函数来检查两个给定的字符串是否是彼此的字谜。一个字符串的变位词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”是彼此的字谜。

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

下面是上述思想的实现：

C++

// C++ program to check whether two 
// strings are anagrams of each other
#include 
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;
}

输出：

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。

下面是上述思想的实现：

C++

// C++ program to check if two strings
// are anagrams of each other
#include 
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

输出：

The two strings are anagram of each other

方法3（使用一个数组计算字符）：
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，如果所有计数值都是 0，那么这两个字符串是彼此的字谜。感谢Ace提出此优化建议。

C++

// C++ program to check if two strings
// are anagrams of each other
#include 
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;
}

输出：

The two strings are anagram of each other

时间复杂度： O(n)

有关详细信息，请参阅关于检查两个字符串是否是彼此的字谜的完整文章！