cc150-1.3

gistfile1.java

/*
	 * 1.3 Given two strings, write a method to decide if one is a permutation
	 * of the other. 
	 
	 create a hashtable <Integer,Integer> Key is the char of the
	 * String element, value is the count of char in the string travese first
	 * string to initial the hashtable. use the hashtable to compare the second
	 * string chars if any value become 0 return false finally return true;
	 * 
	 * O(n) with space o(n)
	 */
	public static void main(String[] args) {
		// string no same size
		String str1 = "abcdfs";
		String str2 = "cba";
		System.out.println(isPermutation(str1, str2));
		// String same size but not permutaion
		str1 = "fasfasfsahf;uip";
		str2 = "fasfasfsahfauip";
		System.out.println(isPermutation(str1, str2));
		// two null String
		str1 = null;
		str2 = null;
		System.out.println(isPermutation(str1, str2));
		// two empty String
		str1 = "";
		str2 = "";
		System.out.println(isPermutation(str1, str2));
		// two permutaion String
		str1 = "asdfghjkl;l";
		str2 = "asdfghjkl;l";
		System.out.println(isPermutation(str1, str2));

	}

	private static boolean isPermutation(String str1, String str2) {
		Hashtable charTable = new Hashtable();
		if (str1 == null && str2 == null)
			return false;
		if (str1 == null || str2 == null)
			return false;
		if (str1.length() != str2.length())
			return false;
		if (str1.length() == str2.length() && str1.length() <= 1)
			return false;

		for (int i = 0; i < str1.length(); i++) {
			if (charTable.get(str1.charAt(i)) != null) {
				int temp = (int) charTable.get(str1.charAt(i));
				charTable.put(str1.charAt(i), (temp + 1));
			} else {
				charTable.put(str1.charAt(i), 1);
			}
		}
		for (int j = 0; j < str2.length(); j++) {
			if (charTable.get(str2.charAt(j)) != null) {
				int current = (int) charTable.get(str2.charAt(j));
				if (current > 0) {
					charTable.put((str2.charAt(j)), current - 1);
				} else
					return false;

			} else
				return false;
		}

		return true;
	}

}