How to implement the 2048 mini game using js

Recently, a classmate attended a seminar and said that a company asked him to write a 2048 mini-game. After referring to some codes on the Internet, he wrote one himself.

The writing ideas are as follows:
1. Set up the HTML layout. There is nothing much to say about the big box nested inside the small box.
2. Implement game initialization and generate the first two small blocks. Here we need to generate a random 2 or 4 and put it in a random position. In fact, after the layout is completed, you can almost think of using arrays to store numbers. This block mainly uses Math.random() to generate random numbers and random X and Y to insert into the array. Then write a function to update the page, using JS to update the content and CSS style of the small box. Here, this block, the function of initializing the array and generating two random boxes is written as a newGame() function, so that it can be called directly when restarting the game.
3. The most important thing comes, move. When I wrote it myself, I followed the idea of ​​​​moving them first and then considering merging them. It is necessary to leave a variable that can record the position of 0, such as when moving downwards, the code is as follows.

for(var i = 0;i<4;i++){    
    var n=3;    
    for(var j = 3;j>=0;j--){        
    if(board[j][i] == 0){            
    continue;
        }else{            
        if(board[n][i]==0){
                board[n][i] = board[j][i];
                board[j][i] = 0;
            }
            n--;
        }
    }
}

I called the function that continues to generate random boxes at the end of the movement function. At this time, I have to consider a question, how to stop adding boxes when there is actually no movement. I still used a loop to complete this problem. Traverse the loop array, and if there is a 0 in front of the moving direction that is not 0, you can continue to move. For example, to determine whether it can be moved down:

function canMoveDown(){
    for(var i = 0;i<4;i++){        
    for(var j = 3;j>=0;j--){            
    if(board[j][i]!=0&&j!=3){                
    if(board[j+1][i]==0){                    return false;
                }
            }
        }
    }    return true;
}

Then rewrite the previous move function, which was judged earlier. If the return is true, directly return false, and the subsequent move function body will not be executed.
4. After the movement is completed, the issue of merging must be considered. My idea is this. After moving, compare the value of the previous box in the moving direction (if it exists) with the box after the move. If they are equal, set the previous box to 2 times itself and move The last box is set to 0. For example, to move down, modify the code as follows:

for(var i = 0;i<4;i++){    
    var n=3;    
    for(var j = 3;j>=0;j--){        
    if(board[j][i] == 0){            
    continue;
        }else{            
        if(board[n][i]==0){
                board[n][i] = board[j][i];
                board[j][i] = 0;
            }            if(n<3){                
            if(board[n][i] == board[n+1][i]){
                    board[n+1][i] = board[n][i]*2;
                    board[n][i] = 0;                    
                    continue;
                }
            }
            n--;
        }
    }
}

This can only be entered when n<3, excluding the case where it is the bottom box, because it has no boxes to merge. It should be noted here that after merging, the current box becomes 0. As mentioned earlier, n is to help record the position of the 0 box. So it should be noted here that the n value does not need to change, because the current box is 0. If there is still a value To move down, you should move here, otherwise you will find that the upper box will be one space empty from the merged box below, which is different from what you expected. Here we also need to modify the judgment condition of the previous canMoveDown() function. If it can be merged, it must also be judged as being able to move. Modify the if condition to:

if(board[j+1][i]==0||board[j+1][i]==board[j][i]){        
eturn false;}

5. At this point, the main function is basically over. There is also a little bit of judgment about the ending. The end is that 1 has no box and is empty, and 2 cannot move. If no box is empty, iterate through the array to see if all the boxes are not 0. If you can't move, use the canMoveDown() function written earlier and the functions of moving up, moving left, and moving right. If these 5 are true, the game end box will pop up, and then call the new newGame() function to start again. .

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