Data Structure in JAVA

1. String/Array/Matrix

First of all, String in Java is a class that contains a char array and other fields and methods. Without code auto-completion of any IDE, the following methods should be remembered.

toCharArray() //get char array of a String Arrays.sort() //sort an array Arrays.toString(char[] a) //convert to string charAt(int x) //get a char at the specific index length() //string length length //array size substring(int beginIndex) substring(int beginIndex, int endIndex) Integer.valueOf()//string to integer String.valueOf()/integer to string

Strings/arrays are easy to understand, but questions related to them often require advanced algorithm to solve, such as dynamic programming, recursion, etc.

2. Linked List

The implementation of a linked list is pretty simple in Java. Each node has a value and a link to next node.

class Node { int val; Node next;   Node(int x) { val = x; next = null; } }

Two popular applications of linked list are stack and queue.

Stack

class Stack{ Node top;   public Node peek(){ if(top != null){ return top; }   return null; }   public Node pop(){ if(top == null){ return null; }else{ Node temp = new Node(top.val); top = top.next; return temp; } }   public void push(Node n){ if(n != null){ n.next = top; top = n; } } }

Queue

class Queue{ Node first, last;   public void enqueue(Node n){ if(first == null){ first = n; last = first; }else{ last.next = n; last = n; } }   public Node dequeue(){ if(first == null){ return null; }else{ Node temp = new Node(first.val); first = first.next; return temp; } } }

3. Tree & Heap

Tree here is normally binary tree. Each node contains a left node and right node like the following:

class TreeNode{ int value; TreeNode left; TreeNode right; }

Here are some concepts related with trees:

1. Binary Search Tree: for all nodes, left children <= current node <= right children
2. Balanced vs. Unbalanced: In a balanced tree, the depth of the left and right subtrees of every node differ by 1 or less.
3. Full Binary Tree: every node other than the leaves has two children.
4. Perfect Binary Tree: a full binary tree in which all leaves are at the same depth or same level, and in which every parent has two children.
5. Complete Binary Tree: a binary tree in which every level, except possibly the last, is completely filled, and all nodes are as far left as possible

4. Graph

Graph related questions mainly focus on depth first search and breath first search. Depth first search is straightforward, you can just loop through neighbors starting from the root node.

Below is a simple implementation of a graph and breath first search. The key is using a queue to store nodes.

1) Define a GraphNode

class GraphNode{ int val; GraphNode next; GraphNode[] neighbors; boolean visited;   GraphNode(int x) { val = x; }   GraphNode(int x, GraphNode[] n){ val = x; neighbors = n; }   public String toString(){ return "value: "+ this.val; } }

2) Define a Queue

class Queue{ GraphNode first, last;   public void enqueue(GraphNode n){ if(first == null){ first = n; last = first; }else{ last.next = n; last = n; } }   public GraphNode dequeue(){ if(first == null){ return null; }else{ GraphNode temp = new GraphNode(first.val, first.neighbors); first = first.next; return temp; } } }

3) Breath First Search uses a Queue

public class GraphTest {   public static void main(String[] args) { GraphNode n1 = new GraphNode(1); GraphNode n2 = new GraphNode(2); GraphNode n3 = new GraphNode(3); GraphNode n4 = new GraphNode(4); GraphNode n5 = new GraphNode(5);   n1.neighbors = new GraphNode[]{n2,n3,n5}; n2.neighbors = new GraphNode[]{n1,n4}; n3.neighbors = new GraphNode[]{n1,n4,n5}; n4.neighbors = new GraphNode[]{n2,n3,n5}; n5.neighbors = new GraphNode[]{n1,n3,n4};   breathFirstSearch(n1, 5); }   public static void breathFirstSearch(GraphNode root, int x){ if(root.val == x) System.out.println("find in root");   Queue queue = new Queue(); root.visited = true; queue.enqueue(root);   while(queue.first != null){ GraphNode c = (GraphNode) queue.dequeue(); for(GraphNode n: c.neighbors){   if(!n.visited){ System.out.print(n + " "); n.visited = true; if(n.val == x) System.out.println("Find "+n); queue.enqueue(n); } } } } }

Output:

value: 2 value: 3 value: 5 Find value: 5
value: 4

Classic Problems:.....

5. Sorting

Time complexity of different sorting algorithms. You can go to wiki to see basic idea of them.

Algorithm	Average Time	Worst Time	Space
Bubble sort	n^2	n^2	1
Selection sort	n^2	n^2	1
Insertion sort	n^2	n^2
Quick sort	n log(n)	n^2
Merge sort	n log(n)	n log(n)	depends

* BinSort, Radix Sort and CountSort use different set of assumptions than the rest, and so they are not “general” sorting methods. (Thanks to Fidel for pointing this out)

In addition, here are some implementations/demos:

6. Recursion vs. Iteration

Recursion should be a built-in thought for programmers. It can be demonstrated by a simple example.

Question:

there are n stairs, each time one can climb 1 or 2. How many different ways to climb the stairs?

Step 1: Finding the relationship before n and n-1.

To get n, there are only two ways, one 1-stair from n-1 or 2-stairs from n-2. If f(n) is the number of ways to climb to n, then f(n) = f(n-1) + f(n-2)

Step 2: Make sure the start condition is correct.

f(0) = 0;
f(1) = 1;

public static int f(int n){ if(n <= 2) return n; int x = f(n-1) + f(n-2); return x; }

The time complexity of the recursive method is exponential to n. There are a lot of redundant computations.

f(5)
f(4) + f(3)
f(3) + f(2) + f(2) + f(1)
f(2) + f(1) + f(2) + f(2) + f(1)

It should be straightforward to convert the recursion to iteration.

public static int f(int n) {   if (n <= 2){ return n; }   int first = 1, second = 2; int third = 0;   for (int i = 3; i <= n; i++) { third = first + second; first = second; second = third; }   return third; }

For this example, iteration takes less time. You may also want to check out Recursion vs Iteration.

7. Dynamic Programming

Dynamic programming is a technique for solving problems with the following properties:

1. An instance is solved using the solutions for smaller instances.
2. The solution for a smaller instance might be needed multiple times.
3. The solutions to smaller instances are stored in a table, so that each smaller instance is solved only once.
4. Additional space is used to save time.


The problem of climbing steps perfectly fit those 4 properties. Therefore, it can be solve by using dynamic programming.

public static int[] A = new int[100];   public static int f3(int n) { if (n <= 2) A[n]= n;   if(A[n] > 0) return A[n]; else A[n] = f3(n-1) + f3(n-2);//store results so only calculate once! return A[n]; }

Classic problems:

8. Bit Manipulation

Bit operators:

OR (|)	AND (&)	XOR (^)	Left Shift (<<)	Right Shift (>>)	Not (~)
1|0=1	1&0=0	1^0=1	0010<<2 td="">	1100>>2=0011	~1=0

Get bit i for a give number n. (i count from 0 and starts from right)

public static boolean getBit(int num, int i){ int result = num & (1<<i);   if(result == 0){ return false; }else{ return true; } }

For example, get second bit of number 10.

i=1, n=10
1<<1 10="" br="" style="margin: 0px; padding: 0px;">1010&10=10
10 is not 0, so return true;

Classic Problems:
1) Find Single Number
2) Maximum Binary Gap

9. Probability

Solving probability related questions normally requires formatting the problem well. Here is just a simple example of such kind of problems.

There are 50 people in a room, what’s the probability that two people have the same birthday? (Ignoring the fact of leap year, i.e., 365 day every year)

Very often calculating probability of something can be converted to calculate the opposite. In this example, we can calculate the probability that all people have unique birthdays. That is: 365/365 * 364/365 * 363/365 * … * 365-n/365 * … * 365-49/365. And the probability that at least two people have the same birthday would be 1 – this value.

public static double caculateProbability(int n){ double x = 1;   for(int i=0; i<n; i++){ x *= (365.0-i)/365.0; }   double pro = Math.round((1-x) * 100); return pro/100; }

calculateProbability(50) = 0.97

10. Combinations and Permutations

The difference between combination and permutation is whether order matters.

Example 1:

Given 5 numbers – 1, 2, 3, 4 and 5, print out different sequence of the 5 numbers. 4 can not be the third one, 3 and 5 can not be adjacent. How many different combinations?

Example 2:

Given 5 banaba, 4 pear, and 3 apple, assuming one kind of fruit are the same, how many different combinations?

Class Problems: