javascript - 在javascript中比较二维数组和一维数组

Question

var player1=["t1", "t9", "t7", "t8", "t2"];
var player2=["t5", "t3", "t4", "t6"];
var winmoves=[[t1,t2,t3],[t4,t5,t6],[t7,t8,t9],[t1,t4,t7],[t2,t5,t8],[t3,t6,t9],[t1,t5,t9],[t3,t5,t7]];

    else if (moves>=3 && moves<=9) {
                    moves+=1;
                    if (turn==1) {
                        var i=0;
                        turn=2;
                        x.innerHTML="<img src='x.png'/>";
                        player1.push(x.id);
                        while(i<=moves){
                            if (winmoves[i] in player1) {
                                alert("player1 wins");
                                document.getElementById("player1").innerHTML=1;
                                }
                            i+=1;
                        }
                    }

我在javascript中有1d数组player1和2d数组winmoves，我想检查w[0]的所有值是否都存在于p中，等等w[1]，w[2]等。if (player1 中的 winmoves[i]) 条件不起作用。我不知道我是否正在写这篇文章。帮帮我吧，我被困在这里了，我该怎么办。

即使我进行了这些更改，它也不起作用。

else if (moves>=3 && moves<9) {
                        moves+=1;
                        if (turn==1) {
                            var i=0;
                            turn=2;
                            x.innerHTML="<img src='x.png'/>";
                            player1.push(x.id);
                            while(i<=moves){
                                 mapped1 = winmoves.map(a1 => a1.every(e1 => player1.includes(e1)));
                                if (mapped1[i]) {
                                    alert("player1 wins");
                                    document.getElementById("player1").innerHTML=1;
                                    }
                                i+=1;
                            }
                        }
                        else if (turn==2) {
                            turn=1;
                            var i=0;
                            x.innerHTML="<img src='o.png'/>";
                            turn=1;
                            player2.push(x.id);
                            while(i<=moves)
                            { 
                                 mapped2 = winmoves.map(a => a.every(e => player2.includes(e)));
                                if (mapped2[i]) {
                                    alert("player2 wins");
                                    document.getElementById("player2").innerHTML=1;
                                    }
                                i+=1;
                            }   
                        }

                    }

Answer 1

我只需通过 Array.prototype.intersect() 的简单发明来完成此操作，其余的就是如此简单的单行代码。

Array.prototype.intersect = function(a) {
  return this.filter(e => a.includes(e));
};

var player1 = ["t1","t2","t3","t5","t7"],
   winmoves = [["t1","t2","t3"],["t4","t5","t6"],["t7","t8","t9"],["t1","t4","t7"],["t2","t5","t8"],["t3","t6","t9"],["t1","t5","t9"],["t3","t5","t7"]];
   filtered = winmoves.filter(a => a.intersect(player1).length == a.length);
     mapped = winmoves.map(a => a.intersect(player1).length == a.length);
console.log(filtered);
console.log(mapped);

好的，我们有一个通用的数组方法来查找两个数组的交集。 (在它调用的数组和作为参数提供的数组之间)它基本上是一个过滤器，检查第一个数组的每一项以查看它是否包含在第二个数组中。因此我们过滤掉两个数组中都存在的项目。

为了获得过滤后的数组，我们再次使用Array.prototype.filter()。这次我们的第一个数组是 winmoves，其中包含我们将检查与 player1 数组的每个交集的数组。如果交集长度等于 winmove 项目的长度，则意味着 winmove 项目的所有元素都存在于 player1 数组中。因此，我们将该数组项返回到filtered。

根据您的情况，如果不使用 intersect 方法，您可以使用 Array.prototype.every() ，如下所示；

var player1 = ["t1","t2","t3","t5","t7"],
   winmoves = [["t1","t2","t3"],["t4","t5","t6"],["t7","t8","t9"],["t1","t4","t7"],["t2","t5","t8"],["t3","t6","t9"],["t1","t5","t9"],["t3","t5","t7"]];
   filtered = winmoves.filter(a => a.every(e => player1.includes(e)));
     mapped = winmoves.map(a => a.every(e => player1.includes(e)));
console.log(filtered);
console.log(mapped);