Pushori Java編程矩陣

Question

我試圖重新創建pushori益智遊戲，我的主要問題在於下一個問題。每當用戶需要輸入一個令牌時，我們就會把它稱爲「1」，就像下面的例子那樣，它需要以某種方式與當時放置的矩陣行「合併」。

考慮這一點，該令牌是在矩陣意味着用戶需要按下右箭頭鍵做「行合併」爲左側：

0 0 0 0  the 0s are the main game matrix, 0 0 0 0 after you place 
    0 0 0 0  the number 1 is the element  0 0 0 0 it when the user 
    0 0 0 0  that's going to enter in the  0 0 0 0 press the right 
1 0 0 0 0  row (3) from the left direction ->0 0 0 1 arrow key 

這會是基本的例子現在真正的問題在於，當你有下以下：

0 3 0 3  there are many tokens around  0 3 0 3 after you place 
2 1 0 2 0  the matrix and when the user -> 0 2 1 2 it when the user 
    1 0 2 0  wants to input the next token  1 0 2 0 press the right 
    0 1 0 1  now it needs to behave like  0 1 0 1 arrow key 

1 0 2 0行（1）和下一個標記將被輸入的數字2，它也需要它的所有元素合併到新的用戶輸入（數字2）作爲新的結果0 2 1 2行（1），我很困惑，在這部分，因爲我有下一個功能，從目前的矩陣表類左表輸入LeftDirection但我卡住試圖想辦法實際實現此功能

public void enterFromLeftDirection(int index,int actualRow){ 


    where index would be element from the Left Direction that needs to merge 
    and actualRow would be actualRow from the matrix that the user is located 
    however I don't get how I would be able to merge everything as I've already said 


} 

任何提示/建議將不勝感激！提前致謝！

Pushori < - 你可以從我從遊戲中收集來看看遊戲從這裏

Answer 1

因此，該行轉移最簡單的方法是......嗯，這很難簡單介紹一下 - 也許一些代碼將有所幫助：

// We will assume that grid is the name of the matrix in the Table object 

public void enterFromLeft(int numToEnter, int row){ 

    for(int i = grid[row].length - 1; i > 0; i--){ 

     // Now we check if that spot is a 0 - if it is, 
     // we need to shift another number over to fill the space 

     if(grid[row][i] == 0){ 

      // Shifts the rightmost number to this spot by 
      // iterating from right to left over the row to find a non-0 number 

      for(int j = i; j > 0; j--){ 

       // Checks if the number is non-0 - and if it is, sets the 
       // empty space to said number and sets the number to 0 - 
       // This essentially moves the rightmost non-0 tile to the right. 

       if(grid[row][j] != 0){ 

        grid[row][i] = grid[row][j]; 
        grid[row][j] = 0; 

        j = -1; // Ends the for loop 

       } 

      } 

      // If grid[row][i] is still 0, it means that there were no non-0 
      // numbers in the regular grid, we will need to fill that space 
      // with numToEnter. If we do that, there will be no other numbers 
      // we need to shift right, so we can end the loop. 

      if(grid[row][i] == 0){ 

       grid[row][i] = numToEnter; 

       i = -1; // To end the loop 

      } 

     } 

    } 

} 

我還沒有測試過這個，但我相信它會工作。要從右側而不是左側添加數字，請將for循環更改爲以其他方式迭代（以及數字以退出循環 - 用grid[row].length + 1替換-1）。

如果您有任何疑問，請在下面評論。但現在我需要去玩一些pushori。