Hallo zusammen,
ich habe von einer Webseite die folgende Code-Snippet eingetippt zum üben. Das Programm bricht sich ab als ein beliebiges Zahl 0-6 eingetragen wird. Das gewünschte Ergebnis wäre es so, dass das Programm geht weiter bis vier O oder X in einer Reihe gesammelt werden. Kann jemand mir eine Hilfsstellung geben?
ich habe von einer Webseite die folgende Code-Snippet eingetippt zum üben. Das Programm bricht sich ab als ein beliebiges Zahl 0-6 eingetragen wird. Das gewünschte Ergebnis wäre es so, dass das Programm geht weiter bis vier O oder X in einer Reihe gesammelt werden. Kann jemand mir eine Hilfsstellung geben?
Java:
package CC4PlayGround;
import java.util.Scanner;
public class JProbs {
public static String[][] createPattern(){ //the "printBoard" method
/**
* the y-axis is set to 15 as graphically there's an extra row to show the _
* at the bottom and you have doubled the cols that show ||| btw each nums.
*/
String[][] f = new String[7][15];
//loop over each row from up to down
for(int i = 0; i < f.length; i++){
//loop over each col from left to right
for(int j = 0; j <f[i].length; j++){
if(j % 2 == 0){
f[i][j] = " | ";
} else {
f[i][j] = " ";
}
//make the lowest row;
if(i == 6){
f[i][j] = " - ";
}
}
}
return f;
}
//print the board just created
public static void printPattern(String[][] f){
for(int i = 0; i < f.length; i++){
for(int j = 0; j < f.length; j++){
System.out.print(f[i][j]);
}
System.out.println();
}
}
//make a basic move
public static void dropPattern1(String f[][]){
//determine what col the user drops the Pat 1 into
System.out.println("Drop a Pat1 at col (0-6)");
//user col nums into a 1-3-5-7-9-11-13 (convert 1-2-3-4-5-6)
Scanner sc = new Scanner(System.in);
int c = 2*sc.nextInt()+1;
System.out.println("user1 entered");
for(int i = 5; i>=0; i--){
if(f[i][c] == " "){
f[i][c] = "O";
//the empty space has been found, break the loop out
break;
}
}
}
//the same step for the other pattern (pat2)
public static void dropPattern2(String f[][]){
System.out.println("Drop a Pat2 at col (0-6)");
Scanner sc = new Scanner(System.in);
int c = 2*sc.nextInt()+1;
System.out.println("user2 entered");
for(int i = 5; i>=0; i--){
if(f[i][c] == " "){
f[i][c] = "X";
//the empty space has been found, break the loop out
break;
}
}
}
//so, check the winner now!
public static String checkWinner(String f[][]){
/*
horizontal winner - it can be found on any row, so loop over each row starting from 0 to 5
*/
for(int i = 0; i < 6; i++){
//it can be found on any col also, loop over each col starting from 0 to 6 also
for(int j = 0; j < 7; j+=2) {
//the increment is 2 as the col is doubled here
if((f[i][j+1] != " ")
&&(f[i][j+3] != " ")
&&(f[i][j+5] != " ")
&&(f[i][j+7] != " ")
&&(f[i][j+1] == f[i][j+3])
&&(f[i][j+3] == f[i][j+5])
&&(f[i][j+5] == f[i][j+7])){
return f[i][j+1];
}
}
}
/*loop over each odd-numbered col by incrementing with 2 then check for consecutive boxes in the same col
that are the same colour
*/
for(int i = 0; i < 15; i+=2){ //if we need to check each odd-numbered col then the range is set to 15
//the y-axis range shall not go over 4 so it's j < 3 (we need 4 patterns vertically anyway)
for(int j = 0; j < 3; j++){
if((f[j][i] != " ")
&&(f[j+1][i] != " ")
&&(f[j+2][i] != " ")
&&(f[j+3][i] != " ")
&&(f[j][i] == f[j+1][i])
&&(f[j+1][i] == f[j+2][i])
&&(f[j+2][i] == f[j+3][i])){
return f[j][i];
}
}
}
/*leftup-rightdown diagonal *
loop over the 3 uppermost rows then go from let to right col-wise
*/
for(int i = 0; i < 6; i++){
/*
the j start from 1 and increment by 2 until it reaches 7
*/
for(int j = 0; j < 9; j+=2) {
if((f[i][j+1] != " ")
&&(f[i+1][j+2] != " ")
&&(f[i+2][j+4] != " ")
&&(f[i+3][j+6] != " ")
&&(f[i][j] == f[i+1][j+2])
&&(f[i+1][j+2] == f[i+2][j+4])
&&(f[i+2][j+4] == f[i+3][j+6])){
return f[i][j];
}
}
}
/*we're just reversing our trajectory like the above method i.e.
we're starting from the rightmost column instead of the leftmost one like we did above
(leftdown-rightup)
*/
for(int i = 0; i < 3; i++){
for(int j = 7; j < 15; j+=2){
if((f[i][j] != " ")
&&(f[i+1][j-2] != " ")
&&(f[i+2][j-4] != " ")
&&(f[i+3][i-6] != " ")
&&(f[i][j] == f[i+1][j-2])
&&(f[i+1][j-2] == f[i+2][j-4])
&&(f[i+2][j-4] == f[i+3][j-6])){
return f[i][j];
}
}
}
//we need to tell the system we could find any winning match (if our search does not fall into any of these four cases)
return null;
}
public static void main(String[] args) {
String[][] f = createPattern();
boolean loop = true;
int count = 0;
printPattern(f);
while(loop){
if(count % 2 == 0){
dropPattern1(f);
} else {
dropPattern2(f);
}
count++;
printPattern(f);
if(checkWinner(f) != null){
if(checkWinner(f) == "1"){
System.out.println("the Pl1 won");
} else if (checkWinner(f) == "0"){
System.out.println("the PL2 won");
}
loop = false; //no winner, then end the game.
}
}
}
}
