WOLOHAHA

public boolean uniqueChar(String s) {
	if (s == null)
		throw new IllegalArgumentException();
	if (s.length() <= 1)
		return true;

	if (s.length() > 256)
		return false;
	boolean[] iFlag = new boolean[256];
	for (int i = 0; i < s.length(); i++) {

WOLOHAHA

public class Main {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Main so = new Main();
		so.reverse("abcdefghijk");
	}

WOLOHAHA

// Ask the interviewer about details: are the characters case sensitive, 
// are the white space and special characters significant?

/**
 * Solution 1
 * Assume the string falls under a certain encoding. 
 * Check if the two strings have the same character count.
 * Use an extra constant
 * size array of integers to store the frequency of each character in str1,
 * for each character in str2, minus the frequency array's value by 1. In

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public class Main {

	/**
	 * @* Runtime & space
   	    * runtime: O(n)
            * space: O(1)  
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Main so = new Main();

WOLOHAHA

public class Main {

	/**
	 * @Runtime & spaces ||runtime: O(n) ||space: O(n)
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Main so = new Main();
		String s = "a";
		System.out.println(so.compress(s));

WOLOHAHA

/**
 * 1.6 Given an image represented by an NxN matrix, 
 * where each pixel in the image is 4 bytes, write a
 * method to rotate the image by 90 degrees. Can you
 * do this in place?
 * @Runtime & spaces 
 * runtime: O(n^2) 
 * space: O(1)
 * Assumption: the rotation is clockwise.
*/

WOLOHAHA

/**
 * 1.7 Write an algorithm such that if 
 * an element in an MxN matrix is 0, its
 * entire row and column are set to 0.
 * 
 * @Runtime & spaces
 * runtime: O(mn)
 * space: O(1)
*/
//Solution: idea got from zhenzhenanan

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public class Main {

	/**
	 * 1.8 Assume you have a method isSubstring which checks if
	 * one word is a substring of another. Given two strings,
	 * s1 and s2, write code to check if s2 is a rotation of s1
	 * using only one call to isSubstring (e.g.,"waterbottle"is
	 * a rotation of"erbottlewat").
	 * 
	 * Solution: just the same as CTCI

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/**
 * 2.1 Write code to remove duplicates from an unsorted linked list.
 * FOLLOW UP
 * How would you solve this problem if a temporary buffer is not allowed?
 * 
 * Solution1:
 * @Runtime & spaces
 * runtime: O(n)
 * space:   O(n)
 * 

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public class Main {

	/**
	 * 2.2 Implement an algorithm to find the k th
	 * to last element of a singly linked list.
	 * 
	 * we have defined k such that passing in k = 1 would return
	 * the last element,k = 2 would return to the second to last
	 * element,and soon.It is equally acceptable to define k such
	 * that k = 0 would return the last element.