I need a class Person to describe persons. Each person has a name and an array consisting of Person-objects, which represent the person's children. The person class has a method getNumberOfDescendants, which returns an integer equal to the total number of descendants of the person, i.e. his children plus grandchildren plus their children etc. Is there a simple way to do this using recursion?
What if you want to count the descendants in a certain generation only? In other words, getNumberOfDescendants(int generation) would return the number of children if generation=1, number of grandchildren if generation=2 etc.
Sure.
public class Person {
private Person[] myChildren;
public int getNumberOfDescendants() {
if (myChildren == null || myChildren.length==0) return 0;
int myDescendants = 0;
for (Person child:myChildren) {
myDescendants += 1; // for this particular child itself
myDescendants += child.getNumberOfDescendants(); //add the child's children, grandchildren, etc.
}
return myDescendants;
}
}
getNumberOfDescendants()
{
int sum = 0;
for (int n=0; n < descendants.length; n++)
{
sum += 1 + descendants[n].getNumberOfDescendants();
}
return sum;
}
Note that the "1 +" is the only place where we're actually increasing the count. That line gets called once for every descendant in the tree.
It's not really recursion, per se, but an instance of the class could sum the results of calling the getNumberOfDescendants() method for each of its children.
Alternatively, you could make this method faster by having each instance notify its parent whenever it got a new descendant (either a new child or from being notified by a child). That way, its count of the number of descendants would always be up-to-date.
public int getNumberDescendents()
{
int nDesc = 0;
if (descendants != null)
{
for (Person p : descendants)
{
nDesc++; // this child
nDesc += p.getNumberDescendents(); // this child's desc
}
}
return nDesc;
}
By the time I wrote the example up, others had posted basically the same thing, so I am kind of redundantly posting.
Related
I am working on a function, countH(), that is supposed to count the amount of times a given number appears in a linked list. For some reason, I cannot get this to work recursively. I have tried a number of different solutions but I guess I can't get something in the correct place. Sorry if I am asking the question poorly, I struggle to understand recursion formatting sometimes.
Here is the function:
public int count(int i) {
return countH(first, i);
}
private int countH(Node front, int i) { // TODO
int cter = 0;
if (front.next==null) {
return 0;
}
if(front.item == i)
cter++;
return countH(front, cter);
}
This is a late version of my code, I'm sure it was a bit better before I messed with it a bunch to try to get it to work
Thanks!
Every recursive implementation consists of two parts:
base case - that represents a simple edge-case for which the outcome is known in advance. For this task, the base case is a situation the given Node is null. Think about it this way: if a head-node is not initialed it will be null and that is the simplest edge-case that your method must be able to handle. And return value for the base case is 0.
recursive case - a part of a solution where recursive calls a made and where the main logic resides. In the recursive case, you need to check the value of a current node. If it matches the target value, then the result returned by the method will be 1 + countH(cur.next, i), otherwise it will be a result of the subsequent recursive call countH(cur.next, i).
Base case is always placed at the beginning of the method, followed by a recursive case.
And when you are writing a recursive part, one of the most important things that you have to keep in mind is which parameters change from one recursive call to another, and which remains the same. In this case, changes only a Node, the target value i remains the same.
public int count(int i) {
return countH(first, i);
}
private int countH(Node cur, int i) { // `front` replaced by `cur`
if (cur == null) { // not cur.next == null (it'll fail with exception if the head-node is null)
return 0;
}
// int cter = 0; // this intermediate variable isn't needed, it could be incremted by 1 at most during the method execution
// if(cur.item == i)
// cter++;
// return countH(cur, cter); // this line contains a mistake - variable `i` has to be passed as a parameter and `cter` must be added to the result returned by a recursive call
return cur.item == i ? 1 + countH(cur.next, i) : countH(cur.next, i);
}
Suggestion
Follow the comments in the code. I've left your original lines in place so that will be easier to compare solutions. Also, always try to come up will reasonable self-explanatory names for variables (as well as methods, classes, etc). For that reason, I renamed the parameter front to cur (short for current), because it's meant to represent any node, not first or any other particular node.
Side note
This statement is called a ternary operator or inline if statement
cur.item == i ? 1 + countH(cur.next, i) : countH(cur.next, i);
And it's just a shorter syntax for the code below:
if (cur.item == i) {
return 1 + countH(cur.next, i);
} else {
return countH(cur.next, i);
}
You could use either of these constructs in your code. The difference is only in syntax, both will get executed in precisely the same way.
In a linked list, you should have one element and from that you get the value and the next element. So your item could look like (I am omitting getters, setters and exception handling):
class Item {
Object value;
Item next;
}
Then your counter for a specific value could look like
int count(Object valueToCount, Item list) {
int result = 0;
if (valueToCount.equals(list.value)) {
result++; // count this value
}
if (value.next != null) {
result += count(valueToCount, value.next) // add the count from remainder of the list
}
return result;
}
public int count(int i) {
return countH(first, i);
}
private int countH(Node front, int i) { // TODO
if(front==null) {
return 0;
}
if (front.item == i) {
return 1 + countH(front.next, i);
} else {
return countH(front.next, i);
}
}
In this problem, if I make the count variable in the second line static, as shown, the kthSmallest() method computes the wrong answer. If the variable is instead made non-static then the correct answer is computed. Non-static methods can use static variables, so why is there a difference?
class Solution {
public static int count = 0;
public int res = 0;
public int kthSmallest(TreeNode root, int k) {
inorder(root,k);
return res;
}
public void inorder(TreeNode root, int k) {
if (root == null) return;
inorder(root.left,k);
count++;
if (count == k) {
res = root.val;
return;
}
inorder(root.right,k);
}
}
I see no reason why the result of a single run of your kthSmallest() method would be affected by whether count is static, but if you perform multiple runs, whether sequentially or in parallel, you will certainly have a problem. count being static means every instance of class Solution shares that variable, which you initialize once to zero, and then only increment. A second run of the method, whether on the same or a different instance of Solution, will continue with the value of count left by the previous run.
Making count non-static partially addresses that issue, by ensuring that every instance of Solution has its own count variable. You still have a problem with performing multiple kthSmallest() computations using the same instance, but you can perform one correct run per instance. If you're testing this via some automated judge then it's plausible that it indeed does create a separate instance for each test case.
But even that is not a complete solution. You still get at most one run per instance, and you're not even sure to get that if an attempt is made to perform two concurrent runs using the same instance. The fundamental problem here is that you are using instance (or class) variables to hold state specific to a single run of the kthSmallest() method.
You ought instead to use local variables of that method, communicated to other methods, if needed, via method arguments and / or return values. For example:
class Solution {
// no class or instance variables at all
public int kthSmallest(TreeNode root, int k) {
// int[1] is the simplest mutable container for an int
int[] result = new int[1];
inorder(root, k, result);
return result[0];
}
// does not need to be public:
// returns the number of nodes traversed (not necessarily the whole subtree)
int inorder(TreeNode root, int k, int[] result) {
if (root == null) {
return 0;
} else {
// nodes traversed in the subtree, plus one for the present node
int count = inorder(root.left, k, result) + 1;
if (count == k) {
result[0] = root.val;
} else {
count += inorder(root.right, k, result);
}
return count;
}
}
}
I would like to count the number of unique objects which are inside a particular Vector. For example, I have a class Person which has Vector<Person> family. I 'm trying to find a way to count the length of the extended family. There could be relations between people such as:-
Mike is related to Pete;
Mike is related to John;
John is related to Amy;
Mike would have an extended family of 4 (which includes himself).
I figured the only way to do this is with recursion and I have something like this so far.
public int getExtendedFamily(Person person) {
// Add the current person if he/she does not exist
if (!(lineage.contains(person))) {
lineage.add(person);
// If the person has no direct family
if (person.getFamily().size() == 0)
return lineage.size();
else {
// Otherwise iterate through the family members
for (Person familyMember : person.getFamily()) {
return lineage.size() + getExtendedFamily(familyMember);
}
}
} else {
// Person already exists...
return lineage.size();
}
return 0;
}
I've created a new vector lineage to keep track of the unique people I come across.
Obviously I know I'm some way off with this code as my lineage.size() isn't right. I just can't get my head around how to structure the recursion, Any guide would be appreciated. Even if it's just pseudocode!
Kind Regards,
Ben
updated code, appears to work
public int getExtendedFamily(Person person) {
// If the person exists just return counter,
//and go back up the call stack;
if ((lineage.contains(person))) {
return counter;
} else {
// Otherwise add the person to the lineage data structure
// and continue
lineage.add(person);
// Loop through current persons family
for (Person family_member_ : person.getFamily()) {
// If the current family member index is not in the lineage
// then increase the counter and recursively call getExtendedFamily,
// to count that persons unique family members.
if (!lineage.contains(family_member_)) {
counter++;
getExtendedFamily(family_member_);
};
}
}
return counter;
}
It would probably just be the vector's size plus one, assuming the relationships are all saved.
If you are getting into generations, where the descendants are not reported in the grandparent, the recursive way to do it would be to return the size of the vector and return if there are no children.
If there is no requirements that the relationships are fully saved, pass a set to the recursive function and save to the set. Return when the vector is 0.
I am working on a school assignment. The objective is to practice GUI's, clone() methods, and using/ modifying existing code. I am trying to write an equals method in the way the instructor desires-- by using a clone of the object, removing items from the bag (returns boolean based on success or failure to remove).
The bag is represented in an array, and should return true in cases such as {1,2,3} and {3,2,1}, ie order does not matter, only the number of each number present in the arrays.
Here is the issue
It works in most cases, however there is a bug in cases where the bags contain numbers as such: {1,1,2} and {1,2,2} and other similar iterations. It is returning true instead of false.
I believe it has something to do with the remove() method we are supposed to use. If i understand it correctly, it is supposed to put the value at the 'end' of the array and decrease the manyItems counter (this is a variable for number of items in the array, because array.length is by default in the constructor 10.)
The code is largely written by another person. We had to import the existing files and write new methods to complete the task we were given. I have all the GUI part done so i will not include that class, only the used methods in the IntArrayBag class.
A second pair of eyes would be helpful. Thanks.
public class IntArrayBag implements Cloneable
{
// Invariant of the IntArrayBag class:
// 1. The number of elements in the bag is in the instance variable
// manyItems, which is no more than data.length.
// 2. For an empty bag, we do not care what is stored in any of data;
// for a non-empty bag, the elements in the bag are stored in data[0]
// through data[manyItems-1], and we don�t care what�s in the
// rest of data.
private int[ ] data;
private int manyItems;
public IntArrayBag( )
{
final int INITIAL_CAPACITY = 10;
manyItems = 0;
data = new int[INITIAL_CAPACITY];
}
public IntArrayBag clone( )
{ // Clone an IntArrayBag object.
IntArrayBag answer;
try
{
answer = (IntArrayBag) super.clone( );
}
catch (CloneNotSupportedException e)
{ // This exception should not occur. But if it does, it would probably
// indicate a programming error that made super.clone unavailable.
// The most common error would be forgetting the "Implements Cloneable"
// clause at the start of this class.
throw new RuntimeException
("This class does not implement Cloneable");
}
answer.data = data.clone( );
return answer;
}
public int size( )
{
return manyItems;
}
public boolean remove(int target)
{
int index; // The location of target in the data array.
// First, set index to the location of target in the data array,
// which could be as small as 0 or as large as manyItems-1; If target
// is not in the array, then index will be set equal to manyItems;
for (index = 0; (index < manyItems) && (target != data[index]); index++)
// No work is needed in the body of this for-loop.
;
if (index == manyItems)
// The target was not found, so nothing is removed.
return false;
else
{ // The target was found at data[index].
// So reduce manyItems by 1 and copy the last element onto data[index].
manyItems--;
data[index] = data[manyItems];
return true;
}
}
//I added extra variables that are not needed to try to increase readability,
//as well as when i was trying to debug the code originally
public boolean equals(Object obj){
if (obj instanceof IntArrayBag){
IntArrayBag canidate = (IntArrayBag) obj; // i know this can be changed, this was required
IntArrayBag canidateTest = (IntArrayBag) canidate.clone(); //this was created
//as a clone because it was otherwise referring to the same memory address
//this caused items to be removed from bags when testing for equality
IntArrayBag test = (IntArrayBag) this.clone();
//fast check to see if the two objects have the same number of items,
//if they dont will return false and skip the item by item checking
if (test.size() != canidateTest.size())
return false;
//the loop will go through every element in the test bag it will
//then remove the value that is present at the first index of the test bag
for (int i = 0; (i < (test.size()) || i < (canidateTest.size())); i++){
int check = test.data[i];
//remove() returns a boolean so if the value is not present in each bag
//then the conditional will be met and the method will return false
boolean test1 = test.remove(check);
boolean test2 = canidateTest.remove(check);
if (test1 != test2)
return false;
}//end for loop
// if the loop goes through every element
//and finds every value was true it will return true
return true;
}//end if
else
return false;
}//end equals
}
I cannot see the big picture, as I havent coded GUIs in Java before, however, as far as comparing 2 int[] arrays, I would sort the arrays before the comparison. This will allow you to eliminate problem cases like the one you stated ( if sorting is possible), then apply something like:
while(array_1[index]==array_2[index] && index<array_1.length)
{index++;}
and find where did the loop break by checking the final value of index
Is it explicitly stated to use clone? You can achieve it easily by overriding the hashCode() for this Object.
You can override the hashCode() for this object as follows:
#Override
public int hashCode() {
final int prime = 5;
int result = 1;
/* Sort Array */
Arrays.sort(this.data);
/* Calculate Hash */
for(int d : this.data) {
result = prime * result + d;
}
/* Return Result */
return result;
}
#Override
public boolean equals(Object obj) {
if (this == obj) return true;
if (obj == null || this.getClass() != obj.getClass()){
return false;
}
return false;
}
If you want to continue using your implementation for equals to compare test and CandidateTest then also you can compute unique hashes and make decision based on the results.
Here is the code snippet:
/* Assuming that you have put size comparison logic on top
and the two objects are of same size */
final int prime = 31;
int testResult = 1;
int candidateTestResult = 1;
for(int i = 0; i < test.size(); i++) {
testResult = prime * testResult + test.data[i];
candidateTestResult = prime * candidateTestResult + candidateTest.data[i];
}
/* Return Result */
return testResult == candidateTestResult;
I believe the problem is in this line:
for (int i = 0; (i < (test.size()) || i < (canidateTest.size())); i++){
The problem here is that test and canidateTest are the clones that you made, and you are removing elements from those bags. And any time you remove an element from the bag, the size will decrease (because you decrease manyItems, and size() returns manyItems). This means you're only going to go through half the array. Suppose the original size is 4. Then, the first time through the loop, i==0 and test.size()==4; the second time, i==0 and test.size()==3; the third time, i==2 and test.size()==2, and you exit the loop. So you don't look at all 4 elements--you only look at 2.
You'll need to decide: do you want to go through the elements of the original array, or the elements of the clone? If you go through the elements of the clone, you actually never need to increment i. You can always look at test.data[0], since once you look at it, you remove it, so you know test.data[0] will be replaced with something else. In fact, you don't need i at all. Just loop until the bag size is 0, or until you determine that the bags aren't equal. On the other hand, if you go through the elements of this.data (i.e. look at this.data[i] or just data[i]), then make sure i goes all the way up to this.size().
(One more small point: the correct spelling is "candidate".)
Maybe you should try SET interface
view this in detail ：http://www.tutorialspoint.com/java/java_set_interface.htm
A set object cannot contains duplicate elements, so it's suitable for your assignment than build your own class.
For example：[1,1,2] and [1,2,2]
you can use this to test whether they are equal
arr1 = {1,1,2}
arr2 = {1,2,2}
Set<Integer> set = new HashSet<Integer>();
for(int i : arr1){//build set of arr1
if(set.contains(i)==false){
set.add(i)
}
}
for(int i:arr2){
if(set.contains(i)==false){
System.out.println('not equal');
break;
}
}
Hope this is helpful.
I have several vectors of different elements but all extending a class which has a specific function, lets say for example
Vector<classone> one;
Vector<classtwo> two;
Vector<classthree> three;
and classone, classtwo and classthree extend Number, and number has two functions:
doThing()
getValue()
And what i want is to call doThing in the order of the getValues received from all the vectors.
One cheap solution would be to concatenate all the vectors in a single Vector, sort it by value and iterate to call the function, but that makes me have to create a huge new vector, occupying new ram, and since the doThing will happen 60 times a second, if the vectors become big, it might be an overkill, i dont really want to create a new vector just to sort it, is there any other solution using the already existing vectors?
Its Java btw.
If one, two and three are sorted, you could create an custom iterator that checks for a given set of lists what the smallest value at the current position is and proceed there.
Should look similar to this (not tested):
class MultiListIterator {
List<Number>[] lists;
int[] positions;
MultiListIterator(List<Number>... lists) {
this.lists = lists;
positions = new int[lists.length];
}
boolean hasNext() {
for (int i = 0; i < lists.length; i++) {
if (positions[i] < lists[i].length) return true;
}
return false;
}
Number next() {
int bestIndex = -1;
Number bestNumber = null;
for (int i = 0; i < lists.length; i++) {
var p = positions[i];
if (p >= positions[i].length) continue;
Number n = lists[i].get(p);
if (bestNumber == null || n.getValue() < bestNumber.getValue()) {
bestIndex = i;
bestNumer = n;
}
}
if (bestNumber == null) throw new RuntimeException("next() beyond hasNext()");
positions[bestIndex++];
return bestNumber;
}
}
Usage:
MultiListIterator mli = new MultiListIterator(one, two, three);
while (mli.hasNext()) {
mli.next().doThing();
}
You may want to let MultiListIterator implement Iterator<Number>.
Note that Java already has a built-in class Number. Using the same name for your class might lead to a lot of confusion when you forget to import it somewhere.
Premature optimizations are generally a bad idea.
Try the method that came to mind first: creating a giant Vector1 ArrayList and sorting it. If it turns out to be a performance issue, then you can start trying new things.