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@ -3449,7 +3449,7 @@ dp[i, c] = \max(dp[i-1, c], dp[i-1, c - wgt[i-1]] + val[i-1])
<li><strong>终止条件</strong>：当物品编号越界 <span class="arithmatex">\(i = 0\)</span> 或背包剩余容量为 <span class="arithmatex">\(0\)</span> 时，终止递归并返回价值 <span class="arithmatex">\(0\)</span> 。</li>
<li><strong>剪枝</strong>：若当前物品重量超出背包剩余容量，则只能不放入背包。</li>
</ul>
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<div class="highlight"><span class="filename">knapsack.java</span><pre><span></span><code><a id="__codelineno-0-1" name="__codelineno-0-1" href="#__codelineno-0-1"></a><span class="cm">/* 0-1 背包：暴力搜索 */</span>
@ -3626,7 +3626,7 @@ dp[i, c] = \max(dp[i-1, c], dp[i-1, c - wgt[i-1]] + val[i-1])
<h3 id="_2">方法二：记忆化搜索<a class="headerlink" href="#_2" title="Permanent link">&para;</a></h3>
<p>为了保证重叠子问题只被计算一次，我们借助记忆列表 <code>mem</code> 来记录子问题的解，其中 <code>mem[i][c]</code> 对应 <span class="arithmatex">\(dp[i, c]\)</span> 。</p>
<p>引入记忆化之后，<strong>时间复杂度取决于子问题数量</strong>，也就是 <span class="arithmatex">\(O(n \times cap)\)</span> 。</p>
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<div class="highlight"><span class="filename">knapsack.java</span><pre><span></span><code><a id="__codelineno-12-1" name="__codelineno-12-1" href="#__codelineno-12-1"></a><span class="cm">/* 0-1 背包：记忆化搜索 */</span>
@ -3841,7 +3841,7 @@ dp[i, c] = \max(dp[i-1, c], dp[i-1, c - wgt[i-1]] + val[i-1])
<h3 id="_3">方法三：动态规划<a class="headerlink" href="#_3" title="Permanent link">&para;</a></h3>
<p>动态规划实质上就是在状态转移中填充 <span class="arithmatex">\(dp\)</span> 表的过程，代码如下所示。</p>
<div class="tabbed-set tabbed-alternate" data-tabs="3:12"><input checked="checked" id="__tabbed_3_1" name="__tabbed_3" type="radio" /><input id="__tabbed_3_2" name="__tabbed_3" type="radio" /><input id="__tabbed_3_3" name="__tabbed_3" type="radio" /><input id="__tabbed_3_4" name="__tabbed_3" type="radio" /><input id="__tabbed_3_5" name="__tabbed_3" type="radio" /><input id="__tabbed_3_6" name="__tabbed_3" type="radio" /><input id="__tabbed_3_7" name="__tabbed_3" type="radio" /><input id="__tabbed_3_8" name="__tabbed_3" type="radio" /><input id="__tabbed_3_9" name="__tabbed_3" type="radio" /><input id="__tabbed_3_10" name="__tabbed_3" type="radio" /><input id="__tabbed_3_11" name="__tabbed_3" type="radio" /><input id="__tabbed_3_12" name="__tabbed_3" type="radio" /><div class="tabbed-labels"><label for="__tabbed_3_1">Java</label><label for="__tabbed_3_2">C++</label><label for="__tabbed_3_3">Python</label><label for="__tabbed_3_4">Go</label><label for="__tabbed_3_5">JavaScript</label><label for="__tabbed_3_6">TypeScript</label><label for="__tabbed_3_7">C</label><label for="__tabbed_3_8">C#</label><label for="__tabbed_3_9">Swift</label><label for="__tabbed_3_10">Zig</label><label for="__tabbed_3_11">Dart</label><label for="__tabbed_3_12">Rust</label></div>
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<div class="highlight"><span class="filename">knapsack.java</span><pre><span></span><code><a id="__codelineno-24-1" name="__codelineno-24-1" href="#__codelineno-24-1"></a><span class="cm">/* 0-1 背包：动态规划 */</span>
@ -4114,7 +4114,7 @@ dp[i, c] = \max(dp[i-1, c], dp[i-1, c - wgt[i-1]] + val[i-1])
</div>
</div>
<p>在代码实现中，我们仅需将数组 <code>dp</code> 的第一维 <span class="arithmatex">\(i\)</span> 直接删除，并且把内循环更改为倒序遍历即可。</p>
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<div class="highlight"><span class="filename">knapsack.java</span><pre><span></span><code><a id="__codelineno-36-1" name="__codelineno-36-1" href="#__codelineno-36-1"></a><span class="cm">/* 0-1 背包：状态压缩后的动态规划 */</span>