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我正在尝试根据另一个 3D 数组的计算创建一个多维数组。

输入数组的格式为 [shift, start Num, end num]

我应该清楚“，”只是为了视觉目的。我需要每个值在其自己的数组位置，即[value1]=location 0,0 [value2]=location 0,1 [value3]=location 0,2等。

例子：

aInput looks like this
[0,1,5]
[1,1,3]
[2,1,2]

aOutput should look like this
[1,1,1]
[1,1,2]
[1,2,1]
[1,2,2]
[1,3,1]
[1,3,2]
[2,1,1]
[2,1,2]
[2,2,1]
[2,2,2]
[2,3,1]
[2,3,2]
[1,3,2]
[3,1,1]
[3,1,2]
[3,2,1]
[3,2,2]
[3,3,1]
[3,3,2]
[etc]

它需要根据班次增加数组中的项目数。即 0 = shift 为 1 列，shift = 1 为 2 列，shift = 3 为 3 列，依此类推。这是我到目前为止所拥有的，但我不知道如何让它计算任何有变化的东西。

var aInput = new Array();
aInput[0] = new Array("0", "1","5");
aInput[1] = new Array("1", "1","3");
aInput[2] = new Array("2", "1","2");

for (var x=0; x < aInput[x].length; x++){
    //Calculate out number
    if (aInput[x][0] == 0){ // Im sure this should be a loop of some sort, just not sure how to do it
        var i=Number(aInput[x][1]);
        while (i <= Number(aInput[x][2])){
            //Write to output array
            aOutput.push(i);
            i++;
        }
    }
}

在此先感谢您的帮助，我真的很难过这个。

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2 回答 2

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var aOutput = [];
for (var i in aInput) {
    var y = parseInt(aInput[i][1])
    for (var x = y; x <= parseInt(aInput[i][2]); x++) {
        aOutput.push([i, y, x]);
    }
}
于 2013-04-29T23:21:52.120 回答
0 
var aInput = new Array();
aInput[0] = new Array("0", "1", "5");
aInput[1] = new Array("1", "1", "3");
aInput[2] = new Array("2", "1", "2");

var input_indexed = [],
    elem = []; // elem holds the row to be added to the output

// Get aInput[] into a more useful arrangement
for (var i = 0; i < aInput.length; i++) {
    // Save the range of each column
    input_indexed[parseInt(aInput[i][0])] = {
        start: parseInt(aInput[i][1]),
        end: parseInt(aInput[i][2])
    };
    // Initialize elem with the start value of each column
    elem[aInput[i][0]] = parseInt(aInput[i][1]);
}

// Produce the output
aOutput = [];
done = false;
while (!done) {
    aOutput.push(elem.slice(0)); // push a copy of elem into result
    for (i = elem.length - 1;; i--) { // Increment elements from right to left
        if (i == -1) { // We've run out of columns
            done = true;
            break;
        }
        elem[i]++; // Increment the current column
        if (elem[i] <= input_indexed[i].end) {
            // If it doesn't overflow, we're done
            break;
        }
        // When it overflows, return to start value and loop around to next column
        elem[i] = input_indexed[i].start;
    }
}

小提琴

于 2013-04-29T23:51:41.477 回答