按容器高度组织容器内容

Question

我的网站有 2 个容器，每个容器都有一堆相关的内容。将哪些内容放入哪个容器中的组织很重要，但不如容器高度尽可能接近那么重要。

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

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问题：

添加更多项目后您会看到，只是有时一个容器的内容会比另一个容器多得多，但确实会发生。

当这种情况发生时，我想把剩下的物品放在较大的容器中，然后以一种使每个容器的高度尽可能接近的方式将它们分配到容器中。

我该怎么做呢？

Answer 1

我已经发布了使用的理想算法。仍然可以进行一些粗略的计算以使容器看起来整洁。

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);
    }
  }
}

Answer 2

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.

我看到你在这里遇到了类似的问题。假设您的容器为两组，每个盒子的高度为将放入该组中的整数。问题是我们必须以这样一种方式划分整数（盒子），以使两个容器的高度保持最小。

相关链接：

Subset Sum Problem - Wiki

Geeks for Geeks tutorial - 好人