XinboChen

/*
Implement an algorithm to determine if a string has all unique characters what if you can't use additional data structure
Date: 08/26/2014
*/
class solution{
  // Time complexity O(n^2), Space complexity O(1)
  public boolean check( String s ){
    boolean result = true;
    if( s.length() < 2 )// include string is empty
    result = true;

XinboChen

class unique{
  public boolean check(String s){
    if(s.length == 0)
    return true;
    
    String sorted = Sort(s);
    for(int i = 0; i < sorted.length - 1 ; i ++){
      if ( sorted.charAt( i ) == sorted.charAt( i + 1 ))
         return false;
    }

XinboChen

/**Implement a function void reverse(char* str) in C or C++ which reverses a null-terminated string.
Time complexity O(n), space complexity O(1)
Date: 08/27/2014
*/
void swap(char* a, char* b){
  char c = *a;
  *a = *b;
  *b = c;
}

XinboChen

/*1.3 Given two strings, write a method to decide if one is a permutation of the other. 
space complexity:
time complexity:
date: 09/01/2014
*/
public class permutationString{
  public static boolean permutation(String s1, String s2){
    if(s1 ==NULL || s2 == NULL)
     return false;
    if(s1.isEmpty()&& s2.isEmpty())

XinboChen

/*1.4 Write a method to replace all spaces in a string with '%20'. You may assume that the string has sufficient space at the end of the string to hold the additional characters, and that you are given the "true" length of the string. (Note: if implementing in Java, please use a character array so that you can perform this operation in place.)    
EXAMPLE
Input: "Mr John Smith       "
Output: "Mr%20John%20Smith"
Time complexity: O(n)
Space complexity: O(1)
Date: 09/02/2014*/
public class replaceSpace{
  public void replace(char[] str){
    int len = strlen(str);

XinboChen

/*implement a method to perform basic string compression using the 
   counts of repeated characters. For example, the string aabcccccaaa would
   become a2b1c5a3. If the "compressed" string would not become smaller than
   the original string, your method should return the original string.
   Time complexity: O(n)
   Space complexity: O(n)
   Date: 09/02/2014
*/

public class compressed{

XinboChen

public static int[][] setMatrixzero(int[][] matrix){
  int M = matrix.length;
  int N = matrix[0].length;
  boolean setRowZero[] = new boolean [M];
  boolean setColumnZero[] = new boolean[N];
  
  for(int i = 0; i < M; i++){
    for(int j=0; j<N; j++)
    if(matrix[i][j] ==0){
      setRowZero[i]=0;

XinboChen

/*
Date: 09/14/2014
Time Complexity:
Space Complexity:
*/

public class solution{
  public static boolean isSubString(String s1, String s2){
    if(s1.contain(s2))
      return true;

XinboChen

class deleteDup{
  public void deleteDup(LinkNode* head){
    if(head == NULL)
        return;
    LinkNode* curr = head;
    while(curr!=NULL){
      LinkNode runner = curr;
      while(runner.next!=NULL){
        if(runner.next.value == curr.value){
          runner.next = runner.next.next;

XinboChen

/*
2.2 Implement an algorithm to find the kth to last element of a singly linked list.
*/

public class findElement{
  public int findElement(LinkNode* head,int k){
    if(head == NULL)
    return;
    LinkNode* curr = head;
    LinkNode* kStep = head;