Question

我正在用阵列解决3个问题。其中两个我已经解决了，我有两个问题。

如果前两个的代码是好的，以及如何解决第三个问题。

好的，第三个问题是以这种方式将数组复制到另一个数组：arrayA = [1,2,3]，然后arrayB是[1,2,3,3,2,1]。

首先从开始到结束然后从结束开始。第一个问题是以这种方式将arrayA复制到arrayB：arrayA = [1,2,3]，arrayB = [1,1,2,2,3， 3]。我只会发布我的伪代码。

//object constructor
function cell(){
    this.alive = Math.random() > 0.7;   
    this.neighbours = 0;  //number of live neighbours
    this.checkneighbours = [[-1,-1],[-1,0],[0,-1],[-1,1],[1,-1],[1,0],[0,1],[1,1]];
}


function GoL(size){
    this.size = size;
    this.grid = this.makeGrid(size);
};      

GoL.prototype.makeGrid = function(size){
    var grid = [];
    for(var i=0; i<size; i++){
        var row=[];
        for(var j =0; j<size; j++){
            row.push(new cell());   
        }
        grid.push(row);
    }
    return grid;
};  

GoL.prototype.drawGrid = function(){
    for(var i=0;i<this.size;i++){
        var row =this.grid[i];
        var rowCell="";
        for(var j=0;j<this.size;j++){
            var cell = row[j];
            if(cell.alive){
                rowCell += "X|";
            }else{
                rowCell += " |";
            }               
        }
        console.log(rowCell);
    }       
};  

GoL.prototype.underpopulation = function(ro,col){
    var cell = this.grid[ro][col];
    if(cell.neighbours <2){
        return true;
    }else{
        return false;   
    }
};  
GoL.prototype.overpopulation = function(ro,col){
    var cell = this.grid[ro][col];
    if(cell.neighbours >3){
        return true;
    }else{
        return false;   
    }
};  

GoL.prototype.backtolife = function(ro,col){
    var cell = this.grid[ro][col];
    if(cell.neighbours ===3 && !cell.alive){
        return true;
    }else{
        return false;   
    }   
};

GoL.prototype.update = function(ro,col){    
    var cell = this.grid[ro][col];
    cell.num_of_neighbours = 0;
    for(var i =0; i<this.checkneighbours.length; i++){
        var checkneighbour = this.checkneighbours[i];
        var neighbour1 = direction[0];
        var neighbour2 = direction[1];
        if(neighbour1>=0 && neighbour1 < this.size && neighbour2 >=0 && neighbour2 < this.size){
            var currentneighbour = this.grid[ro + neighbour1][col+neighbour2];
            if(currentneighbour.alive){
                cell.num_of_neighbours++;
            }
        }
    }
};

GoL.prototype.updateAll = function(){
    for(var i=0; i<this.size;i++){
        for(var j=0; j<this.size;j++){
            this.update(i,j);
        }
    }   
}

GoL.prototype.cellstatus = function(ro,col){
    var cell = this.grid[ro][col];
    if(this.underpopulation(ro,col) || this.overpopulation(ro,col)){
        cell.alive = false;
    }else if(this.backtolife(ro,col)){
        cell.alive = true;
    }
};

GoL.prototype.allcellstatus = function(ro,col){
    for(var i=0; i<this.size;i++){
        for(var j=0; j<this.size;j++){
            this.cellstatus(i,j);
        }
    }   
};


var gameoflife = new GoL(40);   

var interval = setInterval(function(){
    GoL.drawGrid();
    GoL.updateAll();
    GoL.allcellstatus();
},200);

另一个是将arrayA复制到arrayB两次.arrayA = [1,2,3]，arrayB = [1,2,3,1,2,3]。我在这里用于循环：

while i<length(a) do begin
  b[j]=a[i];
  j+=1;
  if (j+1) MOD 2 =0 then i+=1;
end;

Answer 1

如果预先计算了数量，则避免//112233 for i := 1 to length(a) do begin b[2 * i - 1] := a[i]; b[2 * i] := a[i]; end; //123123 l := length(a); for i := 1 to length(a) do begin b[i] := a[i]; b[i + l] := a[i]; end; //123321 l := length(a); for i := 1 to length(a) do begin b[i] := a[i]; b[2*l - i + 1] := a[i]; end;个周期利用简单的索引算术（我假设所有数组都是基于1的）：

database.fetchAllSubscriptionsWithCompletionHandler({subscriptions, error in

如何以不同的方式将数组复制到另一个数组 - Pascal

1 个答案: