Question

我有一个像下面的对象

Product {

String name;

boolean isFiltered;

}

我们有两个列表 - 包含product的masterProductList和filteredProductList。

我们必须将filteredProductList的值（此列表具有属性isFiltered set）复制到masterProductList。

我可以通过以下方法来实现 -

方法1

for (Product filterProduct: filteredPL) {
    for (Product masterProduct: masterPL) {
        if (filterProduct.equals(masterProduct)) {
            // Copy Values
            masterProduct.setIsFiltered(filterProduct.getIsFiltered());
        }
    }
}

方法2

for (Product filterProduct: filteredPL) {
    for (Product masterProduct: masterPL) {
        //Delete the old product having old values
        masterProduct.remove(filterProduct);
        //Add product with new values
        masterProduct.add(filterProduct);
    }
}

这两种方法都具有 n square 复杂度。

还有其他更好的方法可以降低复杂性吗？

Answer 1

您可以将主要产品列表转换为具有常量时间访问方法的哈希集。

Set<Product> masterPLHS = new HashSet<Product>(masterPL); // O(n)

for (Product filterProduct: filteredPL) { //O(n)
        if (masterPLHS.contains(filterProduct)) {  //O(1)
            // do stuff
        }
    }
}

Answer 2

创建代码：

 Hashmap<String, Boolean> merged =new Hashmap<>();

 for(Product product : masterList)
       Merged.put(product.getName(), product.isfiltered());

 for(Product product : filteredList)
       Merged.put(product.getName(), product.isfiltered());

这是组合列表的解决方案。这里，如果name在两个数组列表中，那么根据您的要求，它在hashmap中可以是唯一的，值为filteredlist。

这里复杂性的一个主要因素是

O（n + n）= O（2n）= O(n)

由于

Answer 3

使用第三个列表怎么样？使用第三个列表会阻止List<Object> listA; List<Object> listB; List<Object> listMerged = new List<>(); // init listMerged listMerged.addAll(listA); for(Object elt: listB){ if (!listA.contains(elt)) listMerged.add(elt); }在搜索中包含添加的元素。

<?php
foreach ($_GET as $gkey => $gvalue) {

if (!is_array($gvalue)) {
    $_GET[$gkey] = stripslashes($gvalue);
}
}
?>
<div class="col-md-7 text-right sm-text-right">
<?php
if ($_COOKIE[userid] > 0) {
    echo widget('Bootstrap Theme - Header - Member                             Links','',$w[website_id],$w);

} else { ?>
    <ul class="mini-nav nomargin list-inline <?php if ($wa[custom_159] != 4) { ?>vpad<?php } ?>">
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        if ($wa[custom_148] == "1") { ?>
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            </li>
            <?php
        }
        echo menuArray("header_mini_nav",0,$w); ?>
    </ul>
    <?php
} ?>
<div class="clearfix"></div>
<?php
if (($wa[custom_159] != 0 || $wa[custom_159] == "") && $_SERVER['HTTP_HOST'] != "www.securemypayment.com") {

    if($wa[custom_159] == 4) { ?>
        [widget=Bootstrap Theme - Banner - 320_50]
        <?php
    } else {
        $featurename = '';

        if ($wa[custom_159] == 3 || $wa[custom_159] == 1 || $wa[custom_159] == 2) {
            $featurename = 'search_results';
        }
        if ($wa[custom_159] == 5) {
            $featurename = 'coupons';
        }
        if ($wa[custom_159] == 6) {
            $featurename = 'events';
        }
        if ($wa[custom_159] == 7) {
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        }
        if ($wa[custom_159] == 8) {
            $featurename = 'products';
        }
        if ($wa[custom_159] == 9) {
            $featurename = 'properties';
        }
        if ($wa[custom_159] == 10) {
            $featurename = 'classifieds';
        }
        if ($wa[custom_159] == 11) {
            $featurename = 'videos';
        }
        if ($wa[custom_159] == 12) {
            $featurename = 'blog';
        }
        if ($wa[custom_159] == 13) {
            $featurename = 'articles';
        }
        ?>
        <form action="/<?php echo $featurename; ?>" name="frm1" class="form-inline website-search">
            <?php
            if($wa[custom_159] == 3 || $wa[custom_159] == 2 || $wa[custom_159] == 5 || $wa[custom_159] == 6 || $wa[custom_159] == 7 || $wa[custom_159] == 8 || $wa[custom_159] == 9 || $wa[custom_159] == 10 || $wa[custom_159] == 11 || $wa[custom_159] == 12 || $wa[custom_159] == 13) { ?>
                <div class="input-group input-group-sm bmargin sm-autosuggest">
                    <span class="input-group-addon hidden-md"><i class="fa fa-search"></i></span>
                    <input type="text" placeholder="<?php if ($wa[custom_159] == 3 || $wa[custom_159] == 1 || $wa[custom_159] == 2) { echo $label[keyword_search_default]; } else { ?>Search by Keyword...<? } ?>" value="<? if ($_GET[q]!='') { echo htmlspecialchars($_GET[q]); } ?>" name="q" class=" <? if ($wa[custom_159] == 3 || $wa[custom_159] == 1 || $wa[custom_159] == 2) { ?>member<? } else if ($wa[custom_159] != 9) { echo $featurename; } else { ?>property<? } ?>_search form-control input-sm" autocomplete="off">
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                <?php
            }
            if ($wa[custom_159] == 3 || $wa[custom_159] == 1 || $wa[custom_159] == 10) { ?>
                <div class="input-group input-group-sm bmargin">
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                    <input type="text" placeholder="<?=$label[location_search_default]?>" value="<? if ($_GET[location_value]!="") { echo htmlspecialchars($_GET[location_value]); } else if ($w[geocode_visitor_default]==1 && $w[geocode]==1 && $_SESSION[vdisplay]!="") { echo htmlspecialchars($_SESSION[vdisplay]); } ?>" id="location_google_maps_header" name="location_value" class="googleSuggest googleLocation form-control">
                </div>
                <?php
            } ?>
            <input type="submit" value="Search" class="btn btn-sm btn_search bmargin xs-btn-block">
        </form>
        <?php
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} ?>
</div>

Answer 4

试试这个

Set<Product> masterPLHS = new HashSet<Product>(masterPL);

for (Product filterProduct: filteredPL) { 
      masterPLHS.add(filterProduct);
}

Answer 5

这也应该有效：

List<Product> merged = filteredProductList.stream().filter(masterProductList::contains).collect(Collectors.toList());

编辑：但由于contains方法

，我认为它是n ^ 2

有效复制或合并包含相同对象的两个ArrayList的方法

5 个答案: