代码：

Question

我的网站有2个容器，每个容器都有一堆相关的内容。组织什么内容在什么容器中是重要的，但不像容器在高度上尽可能接近那样重要。

代码：

<button>Click to add more boxes</button>
<div id="wrapper" align="center">
    <div id="container1">
        <div class="box"></div>
        <div class="box"></div>
        <div class="box"></div>
    </div>
    <div id="container2">
        <div class="box"></div>
        <div class="box"></div>
        <div class="box"></div>
        <div class="box"></div>
    </div>
</div>

Jsfiddle

问题：

您会在添加更多项目后看到，有时只有一个容器的内容明显多于另一个容器，但它确实会发生。

当发生这种情况时，我想把剩下的物品放在较大的容器中，然后将它们分配到容器中，使每个容器的高度尽可能接近彼此。

我怎么能这样做呢？

Answer 1

有一个标准算法Subset Sum Problem可以解决这个问题。

问题说

You have n integers. Divide n integers into two sets such that |Set1| - |Set2| is minimum. where |Set(i)| represents the sum of elements in ith set.

我发现你在这里有类似的问题。假设您的容器为两个集合，每个框高度为整数，该整数将放入该集合中。问题是我们必须以两个容器的高度保持最小的方式划分整数（方框）。

Try this bin

JS代码

var $container1 = $('#container1');
var $container2 = $('#container2');
$('.box').each(function(){
    $(this).height(100 * Math.random());
});
var move = $container1.outerWidth();
$container2.css("margin-left",move + 10);

// Let us try to minimize the height difference
var h1 = $container1.height();
var h2 = $container2.height();
var diff = 20;

var currentDiff = h1-h2;
if (currentDiff > diff) {
  alert('1 is bigger than 2');
  reArrange($container1, $container2, Math.abs(currentDiff), diff);
}
if (currentDiff < -diff) {
   alert('2nd is bigger than 1st');
  reArrange($container2, $container1, Math.abs(currentDiff), diff);
}

// Logic: Send the minium height box to smaller container
function reArrange(largeC, smallC, currentDiff, diff) {
    var minHeightBox = null;
    var minH = 10000000000;

    $('.box', largeC).each(function(){
      if ($(this).height() < minH) 
        minHeightBox = this;
        minH = $(this).height();
    });

    // If transfering this to other reduces the difference, transfer this otherwise returh false
  if (minHeightBox !== null) {
    var newDiff = (largeC.height() - minH) - (smallC.height() + minH);
    newDiff = Math.abs(newDiff);
    console.log(newDiff);
    if (newDiff < currentDiff) {
      smallC.append(minHeightBox);
      currentDiff = largeC.height() - smallC.height();      
      //Recursively call the method
      if (currentDiff > diff) 
        reArrange(largeC, smallC, currentDiff, diff);
    }
  }
}

按容器高度组织容器内容

代码：

Jsfiddle

问题：

2 个答案:

Try this bin