Given two strings, write a method to decide if one is a permutation of the other. CC150 - Arrays and Strings.

Permutation.java

package CC150.ArraysAndString;
/**
 * I misunderstand the defination of permutation. I treat it as reverse.
 * Permutation means anyway, but reversed.  abb is a permutation of bab.  
 * Permutation feature: the count of ever number is same, any combination is valid 
 */
public class Permutation {

	public static void main(String[] args) {
		System.out.println(permutation("fabcdfedfaaa","abacdaedfaff"));
	}
	
	/**
	 * Check if the two strings have identical character counts -- the permutation is not same as I think
	 * 
	 * make ever char as position, and increase 	 
	 */
	public static boolean permutation(String s, String t) {
		if (s.length() != t.length()) {
			return false;
		}
		
		int[] arr = new int[256]; // assume character is ASCII
		for (int i=0; i< s.length(); i++) {
			char value = s.charAt(i);
			arr[value] ++;
		}
		
		for (int i=0; i< t.length(); i++) {
			int value = t.charAt(i);
			if (--arr[value] < 0) {
				return false;
			}
		}
		return true;
	}
	
	/**
	 * Sort two strings and compare(use sort api) 
	 */
	public static boolean permutation2(String s, String t) {
		if (s.length() != t.length()) {
			return false;
		}
		return sort(s).equals(sort(t));
	}
	
	public static String sort(String str) {
		char[] temp = str.toCharArray();
		java.util.Arrays.sort(temp);
		return new String(temp);
	}
	
	
	
	/**
	 * My Solution
	 * s[0] == t[length-1]  s[1] == t[length-1-1] .... 
	 * two pointer, one is on s header, the other one is on t end, make comparation
	 */
	
	public static boolean permutation3 (String s, String t) {
		if (s == null || t == null) {
			return false;
		}
		if (s.length() != t.length()) {
			return false;
		}
		
		for (int i= 0; i< s.length(); i++) {
			if (s.charAt(i) != t.charAt(s.length() -1 - i)) {
				return false;
			}
		}
		return true;
		
	}
	
	
	
}

Hmm, I believe that your solution permutation3 is not quite right, since it is checking the first letter of s and last element of t. So, let's say: s = "dog" and t = "odg", which should return true, but when you run your solution, it will print out it is not permutation.