Question

我有一个案例,即Im提供了一台UL和数量不详的LL。 I m 期望利用联合材料将低收入国家分成2 ULs(I m 采用jQuery)。我能以连线的数量来平均分出红灯,但我想根据每个电离层的高度而分立,这样,两个电离层都接近同一高度。

Any help with this would be appreciated, I don t feel like I m getting anywhere with the approach I started with.

感谢。

EDIT: 第一份联合材料。超文本只是直截了当的UL/LI,每种电离层可能不同。

var $sections = $( div.subsection );

$sections.each(function(){
  var $section = $(this);

  var $list = $section.children( ul );
  var $items = $list.children( li );
  var itemCount = $items.size();
  var leftover = itemCount % 2;
  var itemsPerColumn = Math.floor(itemCount / 2);
  var $newList = $( <ul /> );

  $items.each(function(){
    var $this = $(this);
    var index = $items.index($this);

    if (index >= (itemsPerColumn + leftover)) {
      $this.remove().appendTo($newList);
    }
  });

  $list.after($newList);

  _equalizeListHeights();

  function _equalizeListHeights(){
    var listHeight = $list.height();
    var newListHeight = $newList.height();

    if (listHeight > newListHeight){
      var $lastItem = $list.children( li:last );
      var lastItemHeight = $lastItem.height();

      if (listHeight - lastItemHeight > newListHeight + lastItemHeight){
        $lastItem.remove().prependTo($newList);
        _equalizeListHeights();
      }
    }
  }

});

Answer 1

我认为,我至少可以看到以下做法:

Calculate the total height of all the list items (total), and store all the individual heights
Calculate the height of one list (total / 2)
Determine an algorithm to sum a set of heights to come as as possible to total / 2, without exceeding it.
Put the elements with these heights into the first list, and put the rest into the second

第3步是trick。它与

EDIT

这里是解决贵问题的一种新型力量算法。它是在<代码>window.resize上运行的,因为这将是sil。如果你想看到这一变化,将结果窗口转作后推。

//Sum a jQuery wrapped array
$.fn.sum = function() {
    var total = 0;
    this.each(function() { total += this; });
    return total;
};
//Mask elements with a bitmask
$.fn.mask = function(mask) {
    return this.filter(function(i) {
        return (mask >> i) & 1;
    })
}

//Get the sizes, and sneakily return a jQuery object
var sizes = $( .first li ).map(function() { return $(this).outerHeight() });

var total = sizes.sum();
var maxTotal = total / 2;

var best = {
    total: 0,
    mask: 0
}

for (var subsetMask = 1; subsetMask < (1 << sizes.length); subsetMask++) {
    //Sum all the heights in this subset
    var subsetTotal = sizes.mask(subsetMask).sum();

    //New optimal solution?
    if (subsetTotal > best.total && subsetTotal <= maxTotal) {
        best = {
            total: subsetTotal,
            mask: subsetMask
        };
    }
}

$( .first li ).mask(best.mask).appendTo( .second );

Answer 2

You can do it via CSS:

.double_column_list li {float: left; width: 50%;}

<ul class="double_column_list">
    <li>Awesome Stuff Awesome Stuff Awesome Stuff Awesome Stuff Awesome Stuff</li>
    <li>Awesome Stuff</li>
    <li>Awesome Stuff</li>
    <li>Awesome Stuff</li>
</ul>

To get 3 column, set width: 33.333%, 4 column width: 25% and so on.

当然,如果你把一个<代码>li的高度提升到另一个<代码>li的其余部分能够匹配的地步,就会发现坏处。但是,不能通过联合材料确定这一问题。

http://jsfiddle.net/rQJQb/

Update: As pointed out by commenters, if list items are not sorted by height (i.e. height of any one list item in the middle may be bigger/smaller than the ones preceding it), a sorting is needed: http://jsfiddle.net/rQJQb/2/

Answer 3

www.un.org/Depts/DGACM/index_spanish.htm 欧安会问题,你正在努力解决。

You should look into the Backpack Problem. The items to be inserted into the backpack will be the LIs. Each LI will have a weight and value equivalent to its height. The backpack capacity should be half the sum of all the LI heights. Run an algorithm to solve the backpack problem and your LIs will be divided into two sets with heights as you ve described.

<>说明>

背包算法找到了一组物品,其价值尽可能大,但重量不超过背包能力。但是,我们每个低收入国家的重量和价值是高的,后包能力是所有低收入国家合计总高的一半。

因此,从本质上来说,它会给我们带来一系列的全部生命线,这样,最高点尽可能高,不超过总高度1/2。如果你最终拥有两套平等生活设施,这套设备就是其中之一。如果你最后用两套不同高度的低收入人口组成,则后包装问题的解决办法将随着较小的高度而回到定点上(其余部分是非同学的劳动力)。

Solution

Try thecode used here: http://rosettacode.org/wiki/Knapsack_problem/Bounded#Javagust/a>

或者,如果你想推出你自己的话(不是建议的话):

Answer 4

$("ul.first li").each(function() {
    if ($("ul.first").outerHeight() > $("ul.second").outerHeight()) {
        $(this).appendTo($("ul.second"));
    }
});

采用这一分类法。

EDIT

友情链接