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@@ -3541,7 +3541,7 @@ dp[i, c] = \max(dp[i-1, c], dp[i, c - wgt[i-1]] + val[i-1])
\]</div>
<h3 id="_1">代码实现<a class="headerlink" href="#_1" title="Permanent link">&para;</a></h3>
<p>对比两道题目的代码，状态转移中有一处从 <span class="arithmatex">\(i-1\)</span> 变为 <span class="arithmatex">\(i\)</span> ，其余完全一致。</p>
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<div class="highlight"><span class="filename">unbounded_knapsack.java</span><pre><span></span><code><a id="__codelineno-0-1" name="__codelineno-0-1" href="#__codelineno-0-1"></a><span class="cm">/* 完全背包：动态规划 */</span>
@@ -3762,7 +3762,7 @@ dp[i, c] = \max(dp[i-1, c], dp[i, c - wgt[i-1]] + val[i-1])
</div>
</div>
<p>代码实现比较简单，仅需将数组 <code>dp</code> 的第一维删除。</p>
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<div class="highlight"><span class="filename">unbounded_knapsack.java</span><pre><span></span><code><a id="__codelineno-12-1" name="__codelineno-12-1" href="#__codelineno-12-1"></a><span class="cm">/* 完全背包：状态压缩后的动态规划 */</span>
@@ -3989,7 +3989,7 @@ dp[i, a] = \min(dp[i-1, a], dp[i, a - coins[i-1]] + 1)
<p>大多数编程语言并未提供 <span class="arithmatex">\(+ \infty\)</span> 变量，只能使用整型 <code>int</code> 的最大值来代替。而这又会导致大数越界：状态转移方程中的 <span class="arithmatex">\(+ 1\)</span> 操作可能发生溢出。</p>
<p>为此，我们采用数字 <span class="arithmatex">\(amt + 1\)</span> 来表示无效解，因为凑出 <span class="arithmatex">\(amt\)</span> 的硬币个数最多为 <span class="arithmatex">\(amt\)</span> 个。</p>
<p>最后返回前，判断 <span class="arithmatex">\(dp[n, amt]\)</span> 是否等于 <span class="arithmatex">\(amt + 1\)</span> ，若是则返回 <span class="arithmatex">\(-1\)</span> ，代表无法凑出目标金额。</p>
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<div class="highlight"><span class="filename">coin_change.java</span><pre><span></span><code><a id="__codelineno-24-1" name="__codelineno-24-1" href="#__codelineno-24-1"></a><span class="cm">/* 零钱兑换：动态规划 */</span>
@@ -4282,7 +4282,7 @@ dp[i, a] = \min(dp[i-1, a], dp[i, a - coins[i-1]] + 1)
</div>
<h3 id="_4">状态压缩<a class="headerlink" href="#_4" title="Permanent link">&para;</a></h3>
<p>零钱兑换的状态压缩的处理方式和完全背包一致。</p>
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<div class="highlight"><span class="filename">coin_change.java</span><pre><span></span><code><a id="__codelineno-36-1" name="__codelineno-36-1" href="#__codelineno-36-1"></a><span class="cm">/* 零钱兑换：状态压缩后的动态规划 */</span>
@@ -4520,7 +4520,7 @@ dp[i, a] = dp[i-1, a] + dp[i, a - coins[i-1]]
\]</div>
<p>当目标金额为 <span class="arithmatex">\(0\)</span> 时，无需选择任何硬币即可凑出目标金额，因此应将首列所有 <span class="arithmatex">\(dp[i, 0]\)</span> 都初始化为 <span class="arithmatex">\(1\)</span> 。当无硬币时，无法凑出任何 <span class="arithmatex">\(&gt;0\)</span> 的目标金额，因此首行所有 <span class="arithmatex">\(dp[0, a]\)</span> 都等于 <span class="arithmatex">\(0\)</span> 。</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">coin_change_ii.java</span><pre><span></span><code><a id="__codelineno-48-1" name="__codelineno-48-1" href="#__codelineno-48-1"></a><span class="cm">/* 零钱兑换 II：动态规划 */</span>
@@ -4748,7 +4748,7 @@ dp[i, a] = dp[i-1, a] + dp[i, a - coins[i-1]]
</div>
<h3 id="_6">状态压缩<a class="headerlink" href="#_6" title="Permanent link">&para;</a></h3>
<p>状态压缩处理方式相同，删除硬币维度即可。</p>
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<div class="highlight"><span class="filename">coin_change_ii.java</span><pre><span></span><code><a id="__codelineno-60-1" name="__codelineno-60-1" href="#__codelineno-60-1"></a><span class="cm">/* 零钱兑换 II：状态压缩后的动态规划 */</span>