Question

假设你得到以下数组：

foo = [
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,1,1,1,1,1,0,0],
    [0,0,0,1,0,0,0,1,0,0],
    [0,0,0,1,0,0,0,1,0,0],
    [0,0,0,1,1,1,0,1,0,0],
    [0,0,0,0,0,1,0,1,0,0],
    [0,0,0,0,0,1,1,1,0,0],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
]

如何确定1s的模式是否为闭环？几天我一直在努力。我已经尝试了一个递归循环来查找邻居和单词，但是当你有一个更复杂的模式时，它将无法工作，例如：

foo = [
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,1,1,1,0,0,0,0],
    [0,0,0,1,0,1,0,0,0,0],
    [0,0,0,1,0,1,0,0,0,0],
    [0,0,0,1,1,1,1,1,0,0],
    [0,0,0,0,0,1,0,0,0,0],
    [0,0,0,0,0,1,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
]

有人有一个神奇的算法来解决这个问题吗？：（

Answer 1

正如Dagrooms所说，尝试找到1（s）只有一个相邻的1.代码看起来像：

$("#icwsTransferTo").autocomplete({
    minLength: 2,
    source: function(request, response) {

        $.ajax({    
            type: 'GET',
            url: 'index.php',       
            data: {method: 'userSearch', term: request.term},
            dataType: 'json',
            cache: false,
            timeout: 30000,
            success: function(data) { 

            if(!data){
                    return;
                }

                var allFound = $.map(data, function(m) {
                return {
                        'name': m.configurationId.displayName,
                        'ext': m.extension,
                        'id': m.configurationId.id
                    };
                });

                response(allFound);

            }
        }); 

    },
    select: function( event, item) {

        alert(item.value + ' was selected!!!');

    },
    '_renderItem': function (ul, item) {
        return $( "<li>" ).attr( "data-value", item.id)
            .append( '<span>' + item.name+ '</span><span style="float: right"> (' + item.ext + ')</span>' )
            .appendTo( ul );
    }
});

其中行和列是2d数组大小。

<强>更新

如果至少有一个闭环，则返回true：

function isValid1(x,y){
  return (foo[x-1][y] + foo[x+1][y] + foo[x][y-1] + foo[x][y + 1])>1;
}

function validLoop(){
  for(var i = 0; i < rows; i++){
    for(var j = 0; j < columns; j++){
      if(foo[i][j] === 1 && !isValid1(i,j)) {
        return false;
      }
    }
  }
  return true;
}

JSFiddle：https://jsfiddle.net/AdminXVII/b0f7th5d/

更新2 提取循环：

function numTouching1(x,y){
  return foo[x - 1][y] + foo[x + 1][y] + foo[x][y - 1] + foo[x][y + 1];
}

function validLoop(){
  var n = 0, x = 0; // x is current point's number of touching 1 and n is total
  for(var i = 0; i < rows; i++){
    for(var j = 0; j < columns; j++){
      if(foo[i][j] === 1) {
        x = numTouching1(i, j) - 2;
        if(x === -1 || x === 1 || x === 2){
          n += x;
        } 
      }
    }
  }
  return n > -1;
}

JSFiddle：https://jsfiddle.net/AdminXVII/b0f7th5d/7/

更新3

如果有一个以上的循环，这就是威胁，对于一个循环而言，它会变慢。

function numTouching1(x,y){
  return foo[x - 1][y] + foo[x + 1][y] + foo[x][y - 1] + foo[x][y + 1];
}

function extractLoop(){
  for(var i = 0; i < rows; i++){
    for(var j = 0; j < columns; j++){
      if(foo[i][j] === 1 && numTouching1(i, j) === 1){
          foo[i][j] = 0;
          extractLoop();break;
      }
    }
  }
}

JSFiddle：https://jsfiddle.net/AdminXVII/w7zcgpyL/

更新4

更安全function numTouching1(x, y) { return foo[x - 1][y] + foo[x + 1][y] + foo[x][y - 1] + foo[x][y + 1]; } function extractLoop() { for (var i = 0; i < rows; i++) { for (var j = 0; j < columns; j++) { if (foo[i][j] === 1 && numTouching1(i, j) === 1) { foo[i][j] = 0; extractLoop(); break; } } } } function validLoop(){ extractLoop(); for(var i = 0; i < rows; i++){ for(var j = 0; j < columns; j++){ if(foo[i][j] === 1 && numTouching1(i,j) == 2) { return true; } } } return true; }方法：

numTouching1()

修改了之前的JSFiddle

检测2D阵列模式中的闭环

1 个答案: