多维数组 - 检查对角线的连续值 [英] Multi-dimensional Array - Check for diagonal consecutive values

问题描述

写一个数组的检查来确定是否有任何潜在的连续的值在它内部，无论是 horizo​​ntal ， vertical ，或者 diagonal 。下面的例子是一个对角线样例，但是我需要它同时处理 / 和 \ 。

小提琴： http：// jsfiddle .net / PXPn9 / 10 /

所以让我们来假装一下...

  var b = [
 [0,0，X，0，0] 
 [0,0,0，X，0] 
 [0,0,0,0，X] 
 [0,0,0,0] 
 [0,0,0,0] 
] 
  

使用基本的2级深循环，遍历整个事物并使用一些三元运营商识别胜利

  function testWin（）{
 var win = 3，len = b.length，r = 0，c = 0，dr = 0，dl = 0; （var j = 0; j< len; j ++）{
 
 // COL WIN WINCHECK / / 
（b [j] [i] ===X）？ c ++：c = 0; 
 
 // ROW WIN CHECK // 
（b [i] [j] ===X）？ r ++：r = 0; 
 $ b $ // DIAG WIN CHECK // 
 //（b [i] [j] ===X&& b [i + 1] [j + 1] ===X）？ dr ++：dr = 0; 
 //（b [j] [i] ===X&& b [i + 1] [j + 1] ===X）？ dl ++：dl = 0; 
 //赢得所有4 
的检查if（c === win || r === win）{alert（YOU WIN！）;返回true;} 
} 
 r = 0; 
 
 
 
 
 $ 水平检查 code>和垂直检查看起来工作完美，直到我使注释的尝试创建一个对角线测试...我可以有人看看对角线测试，并帮助确定为什么使他们打破了一切，我做了什么错了？
 
 
 我想特别协助创建一个 diagonal 检查。 （查看JSFiddle的全部源代码） 
  // DIAG WIN CHECK // 
 //（b [i] [j] ===X&& b [i + 1] [j + 1] ===X）？ dr ++：dr = 0; 
 //（b [j] [i] ===X&& b [i + 1] [j + 1] ===X）？ dl ++：dl = 0; 
  
 
 
 
 小提琴：  http://jsfiddle.net/PXPn9/10/  
 
 
 
 
 
 
 新的评论 
例如，我试过这个，但它是硬编码为3的对角线（我需要它稍后扩展使用 win 变量）。当我添加这个时，我的水平和垂直检查的右下角失败。
 $ b $（b [i] [j]& b [i + 1] [j + 1]& b [i + 2] [j + 2]）===X）{alert（YOU WON！Diag1）;返回true; }（b [i] [j]&& b [i + 1] [j-1]& b [i + 2] [j-2]）== =X）{alert（YOU WON！Diag2）;返回true; } 
  

我知道这与dl的值有关，dr不能正确复位，它影响到其他的横向和纵向测试，但我有点失去了一个有效的方法来解决这个问题。

解决方案

你试过了：



 （b [i] [j] ===X&& i === j）？ dl ++：0; 
（b [j] [i] ===X&&（i + j）===（len-1））？ dr ++：0; 
  

？

dl 或左对角线将有 i 和 j 相等（so（0， 0）（1,1）和（2,2））

dr 或右对角线将有（0,2）（1,1）等于边长减1的总和 i 和 j （2,0））

当对角线的长度与矩阵的长度相同时，这将起作用，在tic-tac-例如，脚趾游戏。对于部分对角线，您可以稍微修改一下代码：

  var diff = 0; 
 var sum = len  -  1; 
（b [i] [j] ===X）？ diff =（i-j）：diff = 0; 
（b [i] [j] ===X）？ sum =（i + j）：sum = len; 
（b [i] [j] ===X&&（i-j）=== diff）？ dl ++：0; 
（b [j] [i] ===X&&（i + j）=== sum）？ dr ++：0; 
  

Writing a check for an array to determine if there are any potential consecutive values within it, be it horizontal, vertical, or either way diagonal. The example below is a sample diagonal, but I need it working both ways / and \.

Fiddle Away: http://jsfiddle.net/PXPn9/10/

So let's make a pretend scenario...

var b = [ 
    [ 0, 0, X, 0, 0 ]
    [ 0, 0, 0, X, 0 ]
    [ 0, 0, 0, 0, X ]
    [ 0, 0, 0, 0, 0 ]
    [ 0, 0, 0, 0, 0 ]
]

Using a basic 2 level deep loop, which iterates over the whole thing and uses some ternary operators to identify "wins"

function testWin() {
    var win=3, len=b.length, r=0, c=0, dr=0, dl=0;
    for(var i=0;i<len;i++){
        for(var j=0;j<len;j++){

        // COL WIN CHECK //
            (b[j][i]==="X") ? c++ : c=0;

        // ROW WIN CHECK //
            (b[i][j]==="X") ? r++ : r=0;

        // DIAG WIN CHECK //
            // (b[i][j]==="X" && b[i+1][j+1]==="X") ? dr++ : dr=0;
            // (b[j][i]==="X" && b[i+1][j+1]==="X") ? dl++ : dl=0;
        // WIN CHECK FOR ALL 4
            if(c===win || r===win){ alert("YOU WIN!"); return true;}
        }
        r=0;
    }
}

The horizontal check and the vertical check appear to work flawlessly, until I enable the commented attempts to create a diagonal test... Could I have somebody take a look at the diagonal tests and help identify why enabling them breaks everything, and what I have done wrong?

I would like assistance with this in particular to create a diagonal check. (view JSFiddle for whole source)

    // DIAG WIN CHECK //
        // (b[i][j]==="X" && b[i+1][j+1]==="X") ? dr++ : dr=0;
        // (b[j][i]==="X" && b[i+1][j+1]==="X") ? dl++ : dl=0;

Fiddle Away: http://jsfiddle.net/PXPn9/10/

---

NEW COMMENT I've tried this, for example, but it is hard coded for a diagonal of 3 (I need it to expand later to use the win variable). When I add this though, the bottom right corner of my horizontal and vertical checks fails.

    // if((b[i][j] && b[i+1][j+1] && b[i+2][j+2])==="X"){ alert("YOU WON! Diag1"); return true; }
    // if((b[i][j] && b[i+1][j-1] && b[i+2][j-2])==="X"){ alert("YOU WON! Diag2"); return true; }

I know it has something to do with the values of dl and dr aren't resetting properly, and its affecting the other horizontal and vertical tests, but I'm a tad lost of an effective way to solve it.

解决方案

Have you tried:

(b[i][j]==="X" && i===j) ? dl++ : 0;
(b[j][i]==="X" && (i+j)===(len-1)) ? dr++ : 0;

?

dl or the left diagonal would have i and j equal(so (0,0) (1,1) and (2,2))

dr or the right diagonal would have the sum of i and j equal to side-length minus 1(so (0,2) (1,1) (2,0))

This would work when the length of the diagonal is the same as the length of the matrix, full-body diagonal in a tic-tac-toe game, for example. For partial diagonals you can modify the code slightly to something like:

var diff = 0;
var sum = len - 1;  
(b[i][j]==="X") ? diff = (i-j) : diff = 0;  
(b[i][j]==="X") ? sum= (i+j) : sum = len;  
(b[i][j]==="X" && (i-j)===diff) ? dl++ : 0;
(b[j][i]==="X" && (i+j)===sum) ? dr++ : 0;

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