Given two strings, write a method to decide if one is a permutation of the other.

gistfile1.java

/**
 * Solution 1
 * @param str1
 * @param str2
 * @return boolean
 */
public static boolean findIfStringsArePermutationOfEachOther(String str1, String str2) {
    if(str1.length() != str2.length()) {
        return false;
    }
    HashMap<Character, Integer> distinctChars1 = new HashMap<Character, Integer>();
    HashMap<Character, Integer> distinctChars2 = new HashMap<Character, Integer>();
    for(int i = 0; i < str1.length(); i++) {
        Integer charCount1 = distinctChars1.get(str1.charAt(i));
        if(charCount1 == null) {
            charCount1 = 0;
        }
        charCount1++;
        distinctChars1.put(str1.charAt(i), charCount1);

        Integer charCount2 = distinctChars2.get(str2.charAt(i));
        if(charCount2 == null) {
            charCount2 = 0;
        }
        charCount2++;
        distinctChars2.put(str2.charAt(i), charCount2);
    }
    for(Map.Entry<Character, Integer> entry : distinctChars1.entrySet()) {
        if(!distinctChars2.get(entry.getKey()).equals(entry.getValue())) {
            return false;
        }
    }
    return true;
}

/**
 * Solution 2
 * @param s
 * @param t
 * @return boolean
 */
public static boolean permutation(String s, String t) {
    if (s.length() != t.length()) {
        return false;
    }
    // If all chars of string are ascii
    int[] letters = new int[256];
    char[] s_array = s.toCharArray();
    for (char c : s_array) {
        letters[c]++;
    }
    for (int i = 0; i < t.length(); i++) {
        int c = (int) t.charAt(i);
        if (--letters[c] < 0) {
            return false;
        }
    }
    return true;
}