I have the method below, which in my Blackjack app will get the value of the hand which is an NSMutableArray. The problem here is that when 2 Ace's are in a hand, it should be a 12, but because it counts Ace's as 11, it results in being 22, which then makes lowValue returned.
How can I make it so that I can check and see if the for loop has already found an Ace and if it finds another, makes the next Ace worth only 1, not 11?
Thanks!
int getHandValue(NSMutableArray *hand) {
int lowValue = 0;
int highValue = 0;
for (KCCard *aCard in hand) {
if (aCard.value == Ace) {
lowValue+= 1;
highValue+= 11;
} else if (aCard.value == Jack || aCard.value == Queen || aCard.value == King) {
lowValue += 10;
highValue += 10;
} else {
lowValue += aCard.value;
highValue += aCard.value;
}
}
return (highValue > 21) ? lowValue : highValue;
}
Perhaps you can add a boolean value before the for loop setting it initially to NO. When an Ace is found then you can break from the for loop after setting the boolean to YES, that way if you encounter another Ace && your boolean value == YES you can handle the case accordingly.
int getHandValue(NSMutableArray *hand) {
int lowValue = 0;
int highValue = 0;
BOOL isFoundAce = NO;
for (KCCard *aCard in hand) {
if (aCard.value == Ace) {
if (isFoundAce) {
lowValue+= 1;
highValue+= 1;
}
else {
lowValue+= 1;
highValue+= 11;
isFoundAce= YES;
}
} else if (aCard.value == Jack || aCard.value == Queen || aCard.value == King) {
lowValue += 10;
highValue += 10;
} else {
lowValue += aCard.value;
highValue += aCard.value;
}
}
return (highValue > 21) ? lowValue : highValue;
}
My example without a redundant code from zsxwing's example:
int getHandValue(NSMutableArray *hand) {
int cardValue = 0;
int aceCount = 0;
for (KCCard *aCard in hand) {
if (aCard.value == Ace) {
aceCount++;
cardValue += 11;
} else if (aCard.value == Jack || aCard.value == Queen || aCard.value == King) {
cardValue += 10;
} else {
cardValue += aCard.value;
}
}
while ((cardValue > 21) && (aceCount > 0)) {
cardValue -= 10;
aceCount--;
}
return cardValue;
}
Related
Hello I have a connect four game who worked before null safety but I try to make the migration, but I have a problem with scores[i] = null; I can't write like that but without it I have a freeze when CPU is certain to loose
int _compute(Board board, int step, int deepness, List<double> scores) {
for (var i = 0; i < 7; ++i) {
final boardCopy = board.clone();
final target = boardCopy.getColumnTarget(i);
if (target == -1) {
scores[i] = null; // <<<---- HERE I CAN'T USE NULL
continue;
}
final coordinate = Coordinate(i, target);
boardCopy.setBox(coordinate, player);
if (boardCopy.checkWinner(coordinate, player)) {
scores[i] += deepness / (step + 1);
continue;
}
for (var j = 0; j < 7; ++j) {
final target = boardCopy.getColumnTarget(j);
if (target == -1) {
continue;
}
final coordinate = Coordinate(j, target);
boardCopy.setBox(coordinate, otherPlayer);
if (boardCopy.checkWinner(coordinate, otherPlayer)) {
scores[i] -= deepness / (step + 1);
continue;
}
if (step + 1 < deepness) {
_compute(board, step + 1, deepness, scores);
}
}
}
return _getBestScoreIndex(scores);
}
int _getBestScoreIndex(List<double> scores) {
int bestScoreIndex = scores.indexWhere((s) => s != null);
scores.asMap().forEach((index, score) {
if (score != null &&
(score > scores[bestScoreIndex] ||
(score == scores[bestScoreIndex] && _random.nextBool()))) {
bestScoreIndex = index;
}
});
return bestScoreIndex;
}
if I use List<double?>
int _compute(Board board, int step, int deepness, List<double?> scores) {
for (var i = 0; i < 7; ++i) {
final boardCopy = board.clone();
final target = boardCopy.getColumnTarget(i);
if (target == -1) {
scores[i] = null;
continue;
}
final coordinate = Coordinate(i, target);
boardCopy.setBox(coordinate, player);
if (boardCopy.checkWinner(coordinate, player)) {
scores[i] += deepness / (step + 1);//<<<---- HERE I CAN'T USE +=
continue;
}
for (var j = 0; j < 7; ++j) {
final target = boardCopy.getColumnTarget(j);
if (target == -1) {
continue;
}
final coordinate = Coordinate(j, target);
boardCopy.setBox(coordinate, otherPlayer);
if (boardCopy.checkWinner(coordinate, otherPlayer)) {
scores[i] -= deepness / (step + 1); //<<<---- HERE I CAN'T USE -=
continue;
}
if (step + 1 < deepness) {
_compute(board, step + 1, deepness, scores);
}
}
}
return _getBestScoreIndex(scores);
}
int _getBestScoreIndex(List<double?> scores) {
int bestScoreIndex = scores.indexWhere((s) => s != null);
scores.asMap().forEach((index, score) {
if (score != null && // <<<---- HERE I CAN'T USE score !=
(score > scores[bestScoreIndex] || // <<<---- HERE I CAN'T USE score >
(score == scores[bestScoreIndex] && _random.nextBool()))) {
bestScoreIndex = index;
}
});
return bestScoreIndex;
}
The issue with your code is that in the function definition you have defined the data type of score variable as List<double>. Due to which you get an error on assigning score[i] = null. To fix this use data type of score as List<double?>
i believe the param scores is a type-defined value and it has a double, which is a non null value. So you can't assign null to double as it has the type of value assigned.
You can make it double as an optional value, like a List<double?>. but with this, the values in the list will be optional and you need to force unwrap or do null check before using them.
This is a function if the endValueFixed is equal for example 12.0 I want to print the number without zero so I want it to be 12.
void calculateIncrease() {
setState(() {
primerResult = (startingValue * percentage) / 100;
endValue = startingValue + primerResult;
endValueFixe`enter code here`d = roundDouble(endValue, 2);
});
}
This may be an overkill but it works exactly as you wish:
void main() {
// This is your double value
final end = 98.04;
String intPart = "";
String doublePart = "";
int j = 0;
for (int i = 0; i < end.toString().length; i++) {
if (end.toString()[i] != '.') {
intPart += end.toString()[i];
} else {
j = i + 1;
break;
}
}
for (int l = j; l < end.toString().length; l++) {
doublePart += end.toString()[l];
}
if (doublePart[0] == "0" && doublePart[1] != "0") {
print(end);
} else {
print(end.toString());
}
}
You may use this code as a function and send whatever value to end.
if (endValueFixed==12) {
print('${endValueFixed.toInt()}');
}
conditionally cast it to an int and print it then :)
I'm creating a Gmail script that includes 5 variables, one of which is a due date. I just want it to populate as MM/DD/YYYY, however, it is currently populating as Thu Sep 13 2018 00:00:00 GMT-0400 (EDT).
Is there a way I can do that? I've pasted my code below for your reference. Any assistance is much appreciated.
function getRowsData(sheet, range, columnHeadersRowIndex) {
columnHeadersRowIndex = columnHeadersRowIndex || range.getRowIndex() - 1;
var numColumns = range.getEndColumn() - range.getColumn() + 1;
var headersRange = sheet.getRange(columnHeadersRowIndex, range.getColumn(), 1, numColumns);
var headers = headersRange.getValues()[0];
return getObjects(range.getValues(), normalizeHeaders(headers));
}
function getObjects(data, keys) {
var objects = [];
for (var i = 0; i < data.length; ++i) {
var object = {};
var hasData = false;
for (var j = 0; j < data[i].length; ++j) {
var cellData = data[i][j];
if (isCellEmpty(cellData)) {
continue;
}
object[keys[j]] = cellData;
hasData = true;
}
if (hasData) {
objects.push(object);
}
}
return objects;
}
function normalizeHeaders(headers) {
var keys = [];
for (var i = 0; i < headers.length; ++i) {
var key = normalizeHeader(headers[i]);
if (key.length > 0) {
keys.push(key);
}
}
return keys;
}
function normalizeHeader(header) {
var key = "";
var upperCase = false;
for (var i = 0; i < header.length; ++i) {
var letter = header[i];
if (letter == " " && key.length > 0) {
upperCase = true;
continue;
}
if (!isAlnum(letter)) {
continue;
}
if (key.length == 0 && isDigit(letter)) {
continue;
}
if (upperCase) {
upperCase = false;
key += letter.toUpperCase();
} else {
key += letter.toLowerCase();
}
}
return key;
}
function isCellEmpty(cellData) {
return typeof(cellData) == "string" && cellData == "";
}
function isAlnum(char) {
return char >= 'A' && char <= 'Z' ||
char >= 'a' && char <= 'z' ||
isDigit(char);
}
function isDigit(char) {
return char >= '0' && char <= '9';
I don't know how convert java "continue" to scala.
I can use marker from bool + break, but its bad idea
Google did not help :(
It's my first program in scala... yep.. it's horrible
sort java
def sort(in: Array[Int], a:Int, b:Int)
{
var i,j,mode;
double sr=0;
if (a>=b) return; // size =0
for (i=a; i<=b; i++) sr+=in[i];
sr=sr/(b-a+1);
for (i=a, j=b; i <= j;)
{
if (in[i]< sr) { i++; continue; } // left > continue
if (in[j]>=sr) { j--; continue; } // right, continue
int c = in[i]; in[i] = in[j]; in[j]=c;
i++,j--; // swap and continue
}
if (i==a) return;
sort(in,a,j); sort(in,i,b);
}
sort scala...
def SortMerger(in:List[Int], a:Int, b:Int):Unit = {
var i = 0;
var j = 0;
var mode = 0;
var sr = 0.0;
if(a>=b) return;
i=a
while(i<=b)
{
sr+=in.ElementOf(i);
i += 1
}
sr=sr/(b-a+1)
i=a
j=b
while(i<=j)
{
if(in.ElementOf(i)<sr) {
i += 1;
// where continue??? ><
}
}
return
}
Scala has no continue statement, but what you are trying to do can be done with a simple if/else structure.
while(i<=j)
{
if(in(i) < sr) {
i += 1
} else if (in(j) >= sr) {
j -= 1
} else {
int c = in(i)
in(i) = in(j)
in(j) = c
i += 1
j -= 1
}
}
Note that the type of in here should be Array, not List
I believe this can be achieved by counting the instances for each character in that string. Even if a single character in that string is repeated at least twice, we can declare that string as a palindrome.
For example: bbcccc can be rewritten as bccccb or ccbbcc.
edified can be rewritten as deified.
Some book mentioned we should be using hash table. I think we can just use a list and check for the character count.
Do you think the logic is correct?
Yes, the main idea is to count the times of each char existing in the string. And it will be true if the string has at most one char occurs odd times and all others even times.
For example:
aabbcc => acbbca
aabcc => acbca
aabbb => abbba
No. You don't have to use a hash map (as some of the other answers suggest). But the efficiency of the solution will be determined by the algorithm you use.
Here is a solution that only tracks odd characters. If we get 2 odds, we know it can't be a scrambled palindrome. I use an array to track the odd count. I reuse the array index 0 over and over until I find an odd. Then I use array index 1. If I find 2 odds, return false!
Solution without a hash map in javascript:
function isScrambledPalindrome(input) {
// TODO: Add error handling code.
var a = input.split("").sort();
var char, nextChar = "";
var charCount = [ 0 ];
var charIdx = 0;
for ( var i = 0; i < a.length; ++i) {
char = a[i];
nextChar = a[i + 1] || "";
charCount[charIdx]++;
if (char !== nextChar) {
if (charCount[charIdx] % 2 === 1) {
if (charCount.length > 1) {
// A scrambled palindrome can only have 1 odd char count.
return false;
}
charIdx = 1;
charCount.push(0);
} else if (charCount[charIdx] % 2 === 0) {
charCount[charIdx] = 0;
}
}
}
return true;
}
console.log("abc: " + isScrambledPalindrome("abc")); // false
console.log("aabbcd: " + isScrambledPalindrome("aabbcd")); // false
console.log("aabbb: " + isScrambledPalindrome("aabbb")); // true
console.log("a: " + isScrambledPalindrome("a")); // true
Using a hash map, I found a cool way to only track the odd character counts and still determine the answer.
Fun javascript hash map solution:
function isScrambledPalindrome( input ) {
var chars = {};
input.split("").forEach(function(char) {
if (chars[char]) {
delete chars[char]
} else {
chars[char] = "odd" }
});
return (Object.keys(chars).length <= 1);
}
isScrambledPalindrome("aba"); // true
isScrambledPalindrome("abba"); // true
isScrambledPalindrome("abca"); // false
Any string can be palindrome only if at most one character occur odd no. of times and all other characters must occur even number of times.
The following program can be used to check whether a palindrome can be string or not.
vector<int> vec(256,0); //Vector for all ASCII characters present.
for(int i=0;i<s.length();++i)
{
vec[s[i]-'a']++;
}
int odd_count=0,flag=0;
for(int i=0;i<vec.size();++i)
{
if(vec[i]%2!=0)
odd_count++;
if(odd_count>1)
{
flag=1;
cout<<"Can't be palindrome"<<endl;
break;
}
}
if(flag==0)
cout<<"Yes can be palindrome"<<endl;
My code check if can it is palindrome or can be manipulated to Palindrome
#include <stdio.h>
#include <string.h>
#include <stdbool.h>
//Tested on windows 64 bit arhc by using cygwin64 and GCC
bool isPalindrome (char *text);
int main()
{
char text[100]; // it could be N with defining N
bool isPal,isPosPal = false;
printf("Give me a string to test if it is Anagram of Palindrome\n");
gets(text);
isPal = isPalindrome(text);
isPosPal = isAnagramOfPalindrome(text);
if(isPal == false)
{
printf("Not a palindrome.\n");
}
else
{
printf("Palindrome.\n");
}
if(isPosPal == false)
{
printf("Not Anagram of Palindrome\n");
}
else
{
printf("Anagram of Palindrome\n");
}
return 0;
}
bool isPalindrome (char *text) {
int begin, middle, end, length = 0;
length = getLength(text);
end = length - 1;
middle = length/2;
for (begin = 0; begin < middle; begin++)
{
if (text[begin] != text[end])
{
return false;
}
end--;
}
if (begin == middle)
return true;
}
int getLength (char *text) {
int length = 0;
while (text[length] != '\0')
length++;
printf("length: %d\n",length);
return length;
}
int isAnagramOfPalindrome (char *text) {
int length = getLength(text);
int i = 0,j=0;
bool arr[26] = {false};
int counter = 0;
//char string[100]="neveroddoreven";
int a;
for (i = 0; i < length; i++)
{
a = text[i];
a = a-97;
if(arr[a])
{
arr[a] = false;
}
else
{
arr[a] = true;
}
}
for(j = 0; j < 27 ; j++)
{
if (arr[a] == true)
{
counter++;
}
}
printf("counter: %d\n",counter);
if(counter > 1)
{
return false;
}
else if(counter == 1)
{
if(length % 2 == 0)
return false;
else
return true;
}
else if(counter == 0)
{
return true;
}
}
as others have posted, the idea is to have each character occur an even number of times for an even length string, and one character an odd number of times for an odd length string.
The reason the books suggest using a hash table is due to execution time. It is an O(1) operation to insert into / retrieve from a hash map. Yes a list can be used but the execution time will be slightly slower as the sorting of the list will be O(N log N) time.
Pseudo code for a list implementation would be:
sortedList = unsortedList.sort;
bool oddCharFound = false;
//if language does not permit nullable char then initialise
//current char to first element, initialise count to 1 and loop from i=1
currentChar = null;
currentCharCount = 0;
for (int i=0; i <= sortedList.Length; i++) //start from first element go one past end of list
{
if(i == sortedList.Length
|| sortedList[i] != currentChar)
{
if(currentCharCount % 2 = 1)
{
//check if breaks rule
if((sortedList.Length % 2 = 1 && oddCharFound)
|| oddCharFound)
{
return false;
}
else
{
oddCharFound = true;
}
}
if(i!= sortedList.Length)
{
currentCharCount = 1;
currentChar = sortedList[i];
}
}
else
{
currentCharCount++;
}
}
return true;
Here is a simple solution using an array; no sort needed
#include <stdio.h>
#include <stdlib.h>
int main()
{
int a[256] = { 0 };
unsigned char i[] = {"aaBcBccc"};
unsigned char *p = &i[0];
int c = 0;
int j;
int flag = 0;
while (*p != 0)
{
a[*p]++;
p++;
}
for(j=0; j<256; j++)
{
if(a[j] & 1)
{
c++;
if(c > 1)
{
flag = 1;
break;
}
}
}
if(flag)
printf("Nope\n");
else
printf("yup\n");
return 0;
}
C#:
bool ok = s.GroupBy(c => c).Select(g => g.Count()).Where(c => c == 1).Count() < 2;
This solution, however, does use hashing.
Assuming all input characters are lower case letters.
#include<stdio.h>
int main(){
char *str;
char arr[27];
int j;
int a;
j = 0;
printf("Enter the string : ");
scanf("%s", str);
while (*str != '\0'){
a = *str;
a = a%27;
if(arr[a] == *str){
arr[a]=0;
j--;
}else{
arr[a] = *str;
j++;
}
*str++;
}
if(j==0 || j== -1 || j==1){
printf ("\nThe string can be a palindrome\n");
}
}