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POCreation 2D array for Mine game
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  1. POCreation 2D array for Mine game
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    copied!<p><br>Hello <br>I am trying to create emulation of "Mine game" </p> <p><br>suppose we have matrix <br>0 0 0 0 <br>0 0 -1 0 <br>0 0 0 -1 <br>0 -1 0 -1 </p> <p><br>now each cell that not equal -1 should be represent numbers of mines <br>here code for(int i=0;i <pre><code>for(int j=0;j&lt;N;j++) if(box[i][j]!=-1) { switch (i) { case 0: iLEFT=0; iRIGHT=1;break; case 3: iRIGHT=0; iLEFT=1; break; case 1: iLEFT=1; iRIGHT=1; break; case 2: iLEFT=1; iRIGHT=1; break; default: iLEFT=1; iRIGHT=1; break; } switch (j) { case 0: jLEFT=0;` jRIGHT=1; break; case 3: jRIGHT=0; jLEFT=1; break; case 1: jLEFT=1; jRIGHT=1; break; case 2: jLEFT=1; jRIGHT=1; break; default: jLEFT=1; jRIGHT=1; break; } // checking neighbor if(box[i][j+jRIGHT]==-1) count+=1; if(box[i][j-jLEFT]==-1) count+=1; if(box[i+iRIGHT][j]==-1) count+=1; if (box[i-iLEFT][j]==-1) count+=1; if(box[i-iLEFT][j-jLEFT]==-1) { if(i-iLEFT!=i) // trying to avoid double checking count+=1; } if(box[i+iRIGHT][j-jLEFT]==-1) { if(i+iRIGHT!=i) //trying to avoid double checking count+=1; } if (box[i-iLEFT][j+jRIGHT]==-1) { if(i!=iLEFT) //trying to avoid double checking count+=1; } if (box[i+iRIGHT][j+jRIGHT]==-1) { if(i!=iRIGHT) //trying to avoid double checking count+=1; } box[i][j]=count; count=0; } </code></pre> <p><br> My algorithm <br> <br> iLEFT present row step left. <br> iRIGHT present row step right. <br> jLEFT and JRIGHT same for column <br> Suppose i=0 so we can step only one step right(down) <br> if i=1 we can step up and done ..same for j <br></p> <p><br> "Case statement " update iLEFT/iRIGTH and jLEFT/jRIGHT for for enable side steps <br> now "if" statement checking left/right up/done 2 diagonals for box[i][j](only one step always) <br>count counting performance of -1 values neighbor for box[i][j] </p> <p><br><strong>you can see I still have double checking for same cells</strong> <br>0 1 1 1 <br> 0 1 -1 2 <br> 1 2 4 -1 <br> <strong><em>2</em></strong> -1 <strong><em>4</em></strong> -1 </p>
 

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