JamesJi9277

//Implement an algorithm to determine if a string has all unique characters. 
//What if you cannot use additional data structure
//first ask the string is Unicode string or ASCII string
// Unicode is a superset of ASCII, ASCII only defines 128 characters, which mappes 0-127
// but Unicode defines 2^21(maybe less than), and mappes to 0-2^21, 
// but the ASCII can be the same meaning in the Unicode
//the length of string is str.length()
//the length of array is array.length
//array[i], str.charAt(i). array=>[], string => ()
//time complexity is O(n)

JamesJi9277

//Implement a function to reverse a string
// time complexity O(n), space complexity O(n)
public Solution{
	public String reverseString(String str){
		String res = new String();
		if(str == null || str.length() == 0)
			return res;
		for(int i =0; i<str.length();i++)
		{
			res.charAt[i] = str.charAt(str.length() -i);

JamesJi9277

//Given two strings, write a method to decide if one is a permutation of the other
//Note that we will first make a assumption that the comparison is a case sensitive and whitespace is significant. 
// "god     " is different from "dog"
//Solution1, first sort the string, because the string 1 and string 2 are ust characters in different order
//So sort the string first will put the characters from two permutations in the same order
// we just need to compare the sorted verison of two strings
public class Solution1{
	public boolean isPermutation(String str1, String str2){
		if(str1==null||str2==null||str1.length() ==0||str2.length()==0||str1.length()!= str2.length())
			return false;

JamesJi9277

//Write a method to replace all spaces in a string with '%20'
//Bacause the String in Java is immutable, so we will use characters array
public class Solution{
	public void replaceSpace(char[] str, int length){
		if(str == null || str.length == 0)
			return;
		int count =0;
		for(int i = 0;i <str.length;i++)
		{
			if(str[i] == ' ')

JamesJi9277

// Implement a method to perform basic string compression using the counts of repeated characters.
//  For example, the string aabcccccaaa would become a2blc5a3. 
//  If the “compressed” string would not become smaller than the original string, 
//  your method should return the original string.
public class Solution{
	public String compressString(string str){
		StringBuffer res = new StringBuffer();
		char[] str = str.toCharArray();
		char check = str1[0];
		int count = 1;

JamesJi9277

public class Solution{
	public void rotateMatrix(int[][] matrix){
		if(matrix == null || matrix.length == 0|| matrix[0].length ==0)
			return;
		int row = matrix.length;
		for(int i =0; i < row/2; i++)
		{
			for(int j =0;j< n-1-i;j++)
			{
				matrix temp = matrix[i][j];

JamesJi9277

public class Solution{
	public void setZeroes(int[][] matrix){
		if(matrix == null || matrix.length ==0||matrix[0].length == 0)
			return;
		boolean rowHasZero = false;
		boolean colHasZero = false;
		//如果行列里面有0的话，记录一个为0就好，反正都是要变为0的，就先跳出，待程序最后再进行置0的操作，因为第一行和第一列是要被用来作为
		//判断行和判断列的
		for(int i = 0; i< matrix.length; i++)
		{

JamesJi9277

import java.util.Hashtable;
public class Solution1{
	public void deleteDuplicates(LinkedListNode n){
		if(n <= 0)
			return;
		Hashtable<Integer, boolean> res = new Hashtable<Integer, boolean>();
		LinkedListNode prev = null;
		while(n != null)
		{
			if (res.containsKey(n.data))

JamesJi9277

public class Solution{
	public int kthtoLast(LinkedListNode head, int k){
		if (k < 0|| head == null)
			return -1;
		LinkedListNode p1 = head;
		LinkedListNode p2 = head;
		//moving p2 to reach the position length - k.
		for(int i = 0; i< k - 1; i++)
		{
			p2 = p2.next;

JamesJi9277

public class Solution{
	public int kthtoLast(LinkedListNode head, int k){
		if (k < 0|| head == null)
			return -1;
		LinkedListNode p1 = head;
		LinkedListNode p2 = head;
		//moving p2 to reach the position length - k.
		for(int i = 0; i< k - 1; i++)
		{
			p2 = p2.next;