Java 使用HashMap检查两个字符串是否是彼此的变形

写一个函数来检查两个给定的字符串是否是彼此的 嵌合体 。一个字符串的变形是指包含相同字符的另一个字符串，只是字符的顺序可以不同。

例如，”abcd “和 “dabc “是彼此的嵌合。

建议。请先在 {IDE}上尝试你的方法，然后再继续解决。

方法： 哈希图也可以用来寻找任何两个给定的字符串是否是嵌套，方法是将每个字符串的字符映射到单独的哈希图中，然后将它们放在一起比较。

实现

// Java code to check whether two strings
// are Anagram or not using HashMap
 
import java.io.*;
import java.util.*;
 
class GFG {
 
    // Function to check whether two strings
    // are an anagram of each other
    static boolean areAnagram(String str1, String str2)
    {
 
        HashMap<Character, Integer> hmap1
            = new HashMap<Character, Integer>();
        HashMap<Character, Integer> hmap2
            = new HashMap<Character, Integer>();
 
        char arr1[] = str1.toCharArray();
        char arr2[] = str2.toCharArray();
 
        // Mapping first string
        for (int i = 0; i < arr1.length; i++) {
 
            if (hmap1.get(arr1[i]) == null) {
 
                hmap1.put(arr1[i], 1);
            }
            else {
                Integer c = (int)hmap1.get(arr1[i]);
                hmap1.put(arr1[i], ++c);
            }
        }
 
        // Mapping second String
        for (int j = 0; j < arr2.length; j++) {
 
            if (hmap2.get(arr2[j]) == null)
                hmap2.put(arr2[j], 1);
            else {
 
                Integer d = (int)hmap2.get(arr2[j]);
                hmap2.put(arr2[j], ++d);
            }
        }
 
        if (hmap1.equals(hmap2))
            return true;
        else
            return false;
    }
 
    // Test function
    public static void test(String str1, String str2)
    {
 
        System.out.println("Strings to be checked are:\n"
                        + str1 + "\n" + str2 + "\n");
 
        // Find the result
        if (areAnagram(str1, str2))
            System.out.println("The two strings are "
                            + "anagrams of each other\n");
        else
            System.out.println("The two strings are not"
                            + " anagrams of each other\n");
    }
 
    // Driver program
    public static void main(String args[])
    {
 
        // Get the Strings
        String str1 = "geeksforgeeks";
        String str2 = "forgeeksgeeks";
 
        // Test the Strings
        test(str1, str2);
 
        // Get the Strings
        str1 = "geeksforgeeks";
        str2 = "geeks";
 
        // Test the Strings
        test(str1, str2);
    }
}

输出

Strings to be checked are:
geeksforgeeks
forgeeksgeeks

The two strings are anagrams of each other

Strings to be checked are:
geeksforgeeks
geeks

The two strings are not anagram of each other

时间复杂度： O(l1 + l2)，其中l1和l2是字符串的长度。

辅助空间： O(m1+m2)，其中m1和m2是每个字符串中唯一字符的数量。