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<h3 id="_1">参考全排列解法<a class="headerlink" href="#_1" title="Permanent link">&para;</a></h3>
<p>类似于全排列问题，我们可以把子集的生成过程想象成一系列选择的结果，并在选择过程中实时更新“元素和”，当元素和等于 <code>target</code> 时，就将子集记录至结果列表。</p>
<p>而与全排列问题不同的是，<strong>本题集合中的元素可以被无限次选取</strong>，因此无需借助 <code>selected</code> 布尔列表来记录元素是否已被选择。我们可以对全排列代码进行小幅修改，初步得到解题代码。</p>
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<div class="highlight"><span class="filename">subset_sum_i_naive.java</span><pre><span></span><code><a id="__codelineno-0-1" name="__codelineno-0-1" href="#__codelineno-0-1"></a><span class="cm">/* 回溯算法：子集和 I */</span>
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<li>在开启搜索前，先将数组 <code>nums</code> 排序。在遍历所有选择时，<strong>当子集和超过 <code>target</code> 时直接结束循环</strong>，因为后边的元素更大，其子集和都一定会超过 <code>target</code> 。</li>
<li>省去元素和变量 <code>total</code>，<strong>通过在 <code>target</code> 上执行减法来统计元素和</strong>，当 <code>target</code> 等于 <span class="arithmatex">\(0\)</span> 时记录解。</li>
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<div class="highlight"><span class="filename">subset_sum_i.java</span><pre><span></span><code><a id="__codelineno-12-1" name="__codelineno-12-1" href="#__codelineno-12-1"></a><span class="cm">/* 回溯算法：子集和 I */</span>
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<p>为解决此问题，<strong>我们需要限制相等元素在每一轮中只被选择一次</strong>。实现方式比较巧妙：由于数组是已排序的，因此相等元素都是相邻的。这意味着在某轮选择中，若当前元素与其左边元素相等，则说明它已经被选择过，因此直接跳过当前元素。</p>
<p>与此同时，<strong>本题规定中的每个数组元素只能被选择一次</strong>。幸运的是，我们也可以利用变量 <code>start</code> 来满足该约束：当做出选择 <span class="arithmatex">\(x_{i}\)</span> 后，设定下一轮从索引 <span class="arithmatex">\(i + 1\)</span> 开始向后遍历。这样即能去除重复子集，也能避免重复选择元素。</p>
<h3 id="_5">代码实现<a class="headerlink" href="#_5" title="Permanent link">&para;</a></h3>
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<div class="highlight"><span class="filename">subset_sum_ii.java</span><pre><span></span><code><a id="__codelineno-24-1" name="__codelineno-24-1" href="#__codelineno-24-1"></a><span class="cm">/* 回溯算法：子集和 II */</span>