This is my implementation for an insertion sort method using linkedList. I have tried it and it works just fine the only problem that there is an index out of bounds exception caused by the J+1 line. Can anyone tell me how to get around that or how to fix it. Thnx
public static <T extends Comparable <? super T>> void insertionSort2(List<T> portion){
int i = 0;
int j = 0;
T value;
//List <T> sorted = new LinkedList<T>();
// goes through the list
for (i = 1; i < portion.size(); i++) {
// takes each value of the list
value = (T) portion.remove(i);
// the index j takes the value of I and checks the rest of the array
// from the point i
j = i - 1;
while (j >= 0 && (portion.get(j).compareTo(value) >= 0)) {
portion.add(j+1 , portion.remove(j));//it was j+1
j--;
}
// put the value in the correct location.
portion.add(j + 1, value);
}
}
check this code out
just put it as a function in a class and try to call it
void InsertionSort()
{
int temp, out, in;
for(out=1 ; out<size ; out++)
{
temp = list[out];
in = out;
while (in > 0 && list[in-1] > temp)
{
list[in] = list[in-1];
--in;
}
list[in]= temp;
System.out.print("The list in this step became: ");
for (int t=0 ; t<size ; t++)
System.out.print(list[t]+" ");
System.out.println("");
}
}
Related
I m making a card game where 3 random numbers are generated..I need to check are these numbers Row numbers...
like 4 6 5 and 23,24,22. are row numbers
I have made method but I think there should be easy arithmetic formulas
I have tried this and working well, but I need simple arithmatic formula to avoid use of array and for
bool isAllInRow(int num1, int num2,int num3)
{
//subject : tinpati
List<int> numbers=[num1,num2,num3];
bool is_in_row=true;
numbers.sort();
if(numbers[0]==1 && numbers[1]==12 && numbers[2]==13)
return true;
for(int x=0;x<numbers.length-1;x++)
{
if(numbers[x]-numbers[x+1]!=-1)
{
is_in_row=false;
break;
}
}
return is_in_row;
}
So you want to know if the cards form a straight, with aces both low and high.
Is the "three cards" fixed, or would you want to generalize to more cards?
Sorting should be cheap for such a short list, so that's definitely a good start. Then you just need to check the resulting sequence is increasing adjacent values.
I'd do it as:
bool isStraight(List<int> cards) {
var n = cards.length;
if (n < 2) return true;
cards.sort();
var first = cards.first;
if (first == 1 && cards[1] != 2) {
// Pretend Ace is Jack if n == 3.
// Accepts if remaining cards form a straight up to the King.
first = 14 - n;
}
for (var i = 1; i < n; i++) {
if (cards[i] != first + i) return false;
}
return true;
}
This code rejects card sets that have duplicates, or do not form a straight.
I think you are looking for Arithmetic Progression.
bool checkForAP(List<int> numberArr) {
numberArr.sort();
int diff = numberArr[1] - numberArr[0];
if (numberArr[2] - numberArr[1] != diff) {
return false;
}
return true;
}
And modify your function like
bool isAllInRow(int num1, int num2,int num3) {
//subject : tinpati
List<int> numbers=[num1,num2,num3];
bool is_in_row=true;
numbers.sort();
if(numbers[0]==1 && numbers[1]==12 && numbers[2]==13)
return true;
return checkForAP(numbers);
}
Note: remove sort in AP method as it is of no use. Since your numbers
list length is 3 I directly compared numbers for AP, the same can also
be written for n numbers with for.
bool checkForAp(numberArr) {
numberArr.sort();
int diff = numberArr[1] - numberArr[0];
for(int i = 2; i< numberArr.length ;i++) {
if (numberArr[i] - numberArr[i - 1] != diff) {
return false;
}
}
return true;
}
You could do it like this:
bool isAllInRow(int num1, int num2,int num3) {
if (num1 == num2 || num2 == num3) return false;
var maxNum = max(num1, max(num2, num3));
var minNum = min(num1, min(num2, num3));
return (maxNum - minNum == 2) || (minNum == 1 && maxNum == 13 && num1 + num2 + num3 == 26);
}
I want to use the modulo (%) operator in order to access to list items in a circular way. But the code below doesn't work:
void main() {
List<String> myList = ['one', 'two', 'three', 'four'];
int currIdx = 0;
for (int i = 0;i < 10;i++) {
print(myList[currIdx]);
currIdx = currIdx++ % myList.length;
}
}
Here's the solution: instead of coding
currIdx++
you must code
++currIdx
Complete code:
void main() {
List<String> myList = ['one', 'two', 'three', 'four'];
int currIdx = 0;
for (int i = 0;i < 10;i++) {
print(myList[currIdx]);
currIdx = ++currIdx % myList.length; // look second comment below which is very useful !
}
}
I am trying to find the distance of a node from the root of a binary tree but I am getting right answer up to only 3 branches only. like for the node(4) I am getting 3 and for the node (9) and node(10) I am getting 3
#include<bits/stdc++.h>
using namespace std;
struct node
{
int data;
struct node *left;
struct node *right;
node(int val)
{
data = val;
left = NULL;
right = NULL;
}
};
int find_node(node* root,int n)
{
static int length=1;
if (root== NULL)
{
return 0;
}
if (root->data==n)
{
return length;
}
length=length+(find_node(root->left,n)||find_node(root->right,n));
// find_node(root->left,n);
// find_node(root->right,n);
return length;
}
int main ()
{
struct node* root = new node(1);
root->left = new node(2);
root->right = new node(3);
root->left->left = new node(4);
root->left->right = new node(5);
root->right->left = new node(6);
root->right->right = new node(7);
root->right->right->right = new node(9);
root->right->right->right->right = new node(10);
cout <<find_node(root,10);
return 0;}
When your code reaches the first leaf node (with data 4), the following assignment will assign 1:
length=length+(find_node(root->left,n)||find_node(root->right,n));
Because the expression resolves to 1+(0||0), i.e. 1. And so 1 is returned.
The caller (at the node with data 2) will thus receive this 1, and so the above statement will yield 2, since it resolves to 1+(1||......), which is 2 -- the second operand of || is not evaluated.
The parent caller (at the node with data 1), will thus receive this 2. The assignment there resolves to 1+(2||.....), which is again 2 -- realise that || is a logical operator, so it can only evaluate to a boolean value (i.e. 0 or 1).
The issues
In summmary:
You should not use || as it can only evaluate to 0 or 1, losing the actual value from recursion that you need.
You should not use a static variable. For one, it would not reset if you would make a second call to this function from the main program code. Instead, every recursive call should just "mind its own business" and return the depth of n from the given root. The caller should add 1 to that if n was found.
Correction
int find_node(node* root, int n)
{
if (root == NULL)
{
return 0;
}
if (root->data == n)
{
return 1;
}
int length = find_node(root->left, n);
if (!length)
{
length = find_node(root->right, n);
}
if (!length)
{
return 0;
}
return 1 + length;
}
if i'd try to define list in class ,i was facing another problem so i couldn't take list element length as integer.
Hi guys, this is my first stackoverflow question .I hope I could ask well .
I am trying build an calculator like google's But i am facing a problem . When i call buton() method which is recognized in class mantik, everything works right except my list=>List<int> girdi = [];.
Buton method returns kuvvet_alma method then kuvvet_alma assigns a number to list girdi which is recognized out of class .After that kuvvet_alma should proceed on that list his function but isn't.In a forum , a guy have gave me an advice that was using about "breakpoint" but there is no point i've reached .
To sum up i want to add ints in List<int> girdi = []; and works on them with methods like kuvvet_alma or tespit .
Where am i wrong i couldn't get it . I hope i've mentioned well about my problem, thank you everyone who have read it .
List<int> girdi = [];
int n = girdi.length;
class mantik {
void tespit() {
for (int i = n; i >= 0; i--) {
var a = pow(10, n - i);
print(a);
}
}
void kuvvet_alma(int z) {
girdi.add(z);
if (girdi.isNotEmpty == true) {
for (int i = n - 1; i >= 0; i--) {
if (i < 0) {
print(girdi[n]);
} else {
int a = pow(10, i);
print(girdi[(n - 1) - i] * a);
}
}
}
}
void buton(int tus) {
print(girdi);
return mantik().kuvvet_alma(tus);
}
}
for english meaning: girdi=input , mantik=idea, kuvvet_alma=taking power an integer , tespit=determination, buton=button
How can I compute a MD5 or SHA1 hash of text in a specific cell and set it to another cell in Google Spreadsheet?
Is there a formula like =ComputeMD5(A1) or =ComputeSHA1(A1)?
Or is it possible to write custom formula for this? How?
Open Tools > Script Editor then paste the following code:
function MD5 (input) {
var rawHash = Utilities.computeDigest(Utilities.DigestAlgorithm.MD5, input);
var txtHash = '';
for (i = 0; i < rawHash.length; i++) {
var hashVal = rawHash[i];
if (hashVal < 0) {
hashVal += 256;
}
if (hashVal.toString(16).length == 1) {
txtHash += '0';
}
txtHash += hashVal.toString(16);
}
return txtHash;
}
Save the script after that and then use the MD5() function in your spreadsheet while referencing a cell.
This script is based on Utilities.computeDigest() function.
Thanks to gabhubert for the code.
This is the SHA1 version of that code (very simple change)
function GetSHA1(input) {
var rawHash = Utilities.computeDigest(Utilities.DigestAlgorithm.SHA_1, input);
var txtHash = '';
for (j = 0; j <rawHash.length; j++) {
var hashVal = rawHash[j];
if (hashVal < 0)
hashVal += 256;
if (hashVal.toString(16).length == 1)
txtHash += "0";
txtHash += hashVal.toString(16);
}
return txtHash;
}
Ok, got it,
Need to create custom function as explained in
http://code.google.com/googleapps/appsscript/articles/custom_function.html
And then use the apis as explained in
http://code.google.com/googleapps/appsscript/service_utilities.html
I need to handtype the complete function name so that I can see the result in the cell.
Following is the sample of the code that gave base 64 encoded hash of the text
function getBase64EncodedMD5(text)
{
return Utilities.base64Encode( Utilities.computeDigest(Utilities.DigestAlgorithm.MD5, text));
}
The difference between this solution and the others is:
It fixes an issue some of the above solution have with offsetting the output of Utilities.computeDigest (it offsets by 128 instead of 256)
It fixes an issue that causes some other solutions to produce the same hash for different inputs by calling JSON.stringify() on input before passing it to Utilities.computeDigest()
function MD5(input) {
var result = "";
var byteArray = Utilities.computeDigest(Utilities.DigestAlgorithm.MD5, JSON.stringify(input));
for (i=0; i < byteArray.length; i++) {
result += (byteArray[i] + 128).toString(16) + "-";
}
result = result.substring(result, result.length - 1); // remove trailing dash
return result;
}
to get hashes for a range of cells, add this next to gabhubert's function:
function RangeGetMD5Hash(input) {
if (input.map) { // Test whether input is an array.
return input.map(GetMD5Hash); // Recurse over array if so.
} else {
return GetMD5Hash(input)
}
}
and use it in cell this way:
=RangeGetMD5Hash(A5:X25)
It returns range of same dimensions as source one, values will spread down and right from cell with formulae.
It's universal single-value-function to range-func conversion method (ref), and it's way faster than separate formuleas for each cell; in this form, it also works for single cell, so maybe it's worth to rewrite source function this way.
Based on #gabhubert but using array operations to get the hexadecimal representation
function sha(str){
return Utilities
.computeDigest(Utilities.DigestAlgorithm.SHA_1, str) // string to digested array of integers
.map(function(val) {return val<0? val+256 : val}) // correct the offset
.map(function(val) {return ("00" + val.toString(16)).slice(-2)}) // add padding and enconde
.join(''); // join in a single string
}
Using #gabhubert answer, you could do this, if you want to get the results from a whole row. From the script editor.
function GetMD5Hash(value) {
var rawHash = Utilities.computeDigest(Utilities.DigestAlgorithm.MD5, value);
var txtHash = '';
for (j = 0; j <rawHash.length; j++) {
var hashVal = rawHash[j];
if (hashVal < 0)
hashVal += 256;
if (hashVal.toString(16).length == 1)
txtHash += "0";
txtHash += hashVal.toString(16);
}
return txtHash;
}
function straightToText() {
var ss = SpreadsheetApp.getActiveSpreadsheet().getSheets();
var r = 1;
var n_rows = 9999;
var n_cols = 1;
var column = 1;
var sheet = ss[0].getRange(r, column, n_rows, ncols).getValues(); // get first sheet, a1:a9999
var results = [];
for (var i = 0; i < sheet.length; i++) {
var hashmd5= GetMD5Hash(sheet[i][0]);
results.push(hashmd5);
}
var dest_col = 3;
for (var j = 0; j < results.length; j++) {
var row = j+1;
ss[0].getRange(row, dest_col).setValue(results[j]); // write output to c1:c9999 as text
}
}
And then, from the Run menu, just run the function straightToText() so you can get your result, and elude the too many calls to a function error.
I was looking for an option that would provide a shorter result. What do you think about this? It only returns 4 characters. The unfortunate part is that it uses i's and o's which can be confused for L's and 0's respectively; with the right font and in caps it wouldn't matter much.
function getShortMD5Hash(input) {
var rawHash = Utilities.computeDigest(Utilities.DigestAlgorithm.MD5, input);
var txtHash = '';
for (j = 0; j < 16; j += 8) {
hashVal = (rawHash[j] + rawHash[j+1] + rawHash[j+2] + rawHash[j+3]) ^ (rawHash[j+4] + rawHash[j+5] + rawHash[j+6] + rawHash[j+7])
if (hashVal < 0)
hashVal += 1024;
if (hashVal.toString(36).length == 1)
txtHash += "0";
txtHash += hashVal.toString(36);
}
return txtHash.toUpperCase();
}
I needed to get a hash across a range of cells, so I run it like this:
function RangeSHA256(input)
{
return Array.isArray(input) ?
input.map(row => row.map(cell => SHA256(cell))) :
SHA256(input);
}