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<title>13.1. 初探动态规划（New） - Hello 算法</title>
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<title>14.1. 初探动态规划（New） - Hello 算法</title>
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13.1. 初探动态规划（New）
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14.1. 初探动态规划（New）
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12. 分治
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12.1. 分治算法（New）
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12.2. 构建树问题（New）
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<div class="md-nav__link md-nav__link--index ">
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<a href="../../chapter_backtracking/">12. 回溯</a>
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<a href="../../chapter_backtracking/">13. 回溯</a>
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<span class="md-nav__icon md-icon"></span>
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12. 回溯
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13. 回溯
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<li class="md-nav__item">
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<a href="../../chapter_backtracking/backtracking_algorithm/" class="md-nav__link">
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12.1. 回溯算法
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13.1. 回溯算法
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</a>
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<li class="md-nav__item">
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<a href="../../chapter_backtracking/permutations_problem/" class="md-nav__link">
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12.2. 全排列问题
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13.2. 全排列问题
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</a>
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<li class="md-nav__item">
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<a href="../../chapter_backtracking/subset_sum_problem/" class="md-nav__link">
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12.3. 子集和问题
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13.3. 子集和问题
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</a>
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<a href="../../chapter_backtracking/n_queens_problem/" class="md-nav__link">
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12.4. N 皇后问题
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13.4. N 皇后问题
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</a>
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<li class="md-nav__item">
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<a href="../../chapter_backtracking/summary/" class="md-nav__link">
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12.5. 小结
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13.5. 小结
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</a>
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<div class="md-nav__link md-nav__link--index ">
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<a href="../">13. 动态规划</a>
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<a href="../">14. 动态规划</a>
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13. 动态规划
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14. 动态规划
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<label class="md-nav__link md-nav__link--active" for="__toc">
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13.1. 初探动态规划（New）
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14.1. 初探动态规划（New）
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<a href="./" class="md-nav__link md-nav__link--active">
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13.1. 初探动态规划（New）
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14.1. 初探动态规划（New）
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</a>
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<ul class="md-nav__list" data-md-component="toc" data-md-scrollfix>
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<li class="md-nav__item">
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<a href="#1311" class="md-nav__link">
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13.1.1. 方法一：暴力搜索
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<a href="#1411" class="md-nav__link">
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14.1.1. 方法一：暴力搜索
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</a>
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<li class="md-nav__item">
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13.1.2. 方法二：记忆化搜索
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14.1.2. 方法二：记忆化搜索
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</a>
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13.1.3. 方法三：动态规划
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<a href="#1413" class="md-nav__link">
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14.1.3. 方法三：动态规划
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</a>
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<a href="../dp_problem_features/" class="md-nav__link">
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13.2. DP 问题特性（New）
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14.2. DP 问题特性（New）
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</a>
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<li class="md-nav__item">
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<a href="../dp_solution_pipeline/" class="md-nav__link">
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13.3. DP 解题思路（New）
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14.3. DP 解题思路（New）
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</a>
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<a href="../knapsack_problem/" class="md-nav__link">
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13.4. 0-1 背包问题（New）
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14.4. 0-1 背包问题（New）
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</a>
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<a href="../unbounded_knapsack_problem/" class="md-nav__link">
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13.5. 完全背包问题（New）
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14.5. 完全背包问题（New）
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<a href="../edit_distance_problem/" class="md-nav__link">
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13.6. 编辑距离问题（New）
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14.6. 编辑距离问题（New）
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13.7. 小结（New）
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14.1. 编程环境安装
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14.2. 一起参与创作
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14.7. 小结（New）
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15.1. 编程环境安装
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</a>
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参考文献
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</label>
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<ul class="md-nav__list" data-md-component="toc" data-md-scrollfix>
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<a href="#1311" class="md-nav__link">
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13.1.1. 方法一：暴力搜索
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<a href="#1411" class="md-nav__link">
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14.1.1. 方法一：暴力搜索
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</a>
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</li>
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13.1.2. 方法二：记忆化搜索
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14.1.2. 方法二：记忆化搜索
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</a>
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13.1.3. 方法三：动态规划
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<a href="#1413" class="md-nav__link">
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14.1.3. 方法三：动态规划
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<h1 id="131">13.1. 初探动态规划<a class="headerlink" href="#131" title="Permanent link">¶</a></h1>
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<h1 id="141">14.1. 初探动态规划<a class="headerlink" href="#141" title="Permanent link">¶</a></h1>
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<p>「动态规划 Dynamic Programming」是一种通过将复杂问题分解为更简单的子问题的方式来求解问题的方法。它将一个问题分解为一系列更小的子问题，并通过存储子问题的解来避免重复计算，从而大幅提升时间效率。</p>
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<p>在本节中，我们从一个经典例题入手，先给出它的暴力回溯解法，观察其中包含的重叠子问题，再逐步导出更高效的动态规划解法。</p>
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<div class="admonition question">
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</code></pre></div>
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</div>
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<div class="tabbed-block">
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<div class="highlight"><span class="filename">climbing_stairs_backtrack.zig</span><pre><span></span><code><a id="__codelineno-9-1" name="__codelineno-9-1" href="#__codelineno-9-1"></a><span class="p">[</span><span class="n">class</span><span class="p">]{}</span><span class="o">-</span><span class="p">[</span><span class="n">func</span><span class="p">]{</span><span class="n">backtrack</span><span class="p">}</span>
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<a id="__codelineno-9-2" name="__codelineno-9-2" href="#__codelineno-9-2"></a>
|
||||
<a id="__codelineno-9-3" name="__codelineno-9-3" href="#__codelineno-9-3"></a><span class="p">[</span><span class="n">class</span><span class="p">]{}</span><span class="o">-</span><span class="p">[</span><span class="n">func</span><span class="p">]{</span><span class="n">climbingStairsBacktrack</span><span class="p">}</span>
|
||||
<div class="highlight"><span class="filename">climbing_stairs_backtrack.zig</span><pre><span></span><code><a id="__codelineno-9-1" name="__codelineno-9-1" href="#__codelineno-9-1"></a><span class="c1">// 回溯</span>
|
||||
<a id="__codelineno-9-2" name="__codelineno-9-2" href="#__codelineno-9-2"></a><span class="k">fn</span><span class="w"> </span><span class="n">backtrack</span><span class="p">(</span><span class="n">choices</span><span class="o">:</span><span class="w"> </span><span class="p">[]</span><span class="kt">i32</span><span class="p">,</span><span class="w"> </span><span class="n">state</span><span class="o">:</span><span class="w"> </span><span class="kt">i32</span><span class="p">,</span><span class="w"> </span><span class="n">n</span><span class="o">:</span><span class="w"> </span><span class="kt">i32</span><span class="p">,</span><span class="w"> </span><span class="n">res</span><span class="o">:</span><span class="w"> </span><span class="n">std</span><span class="p">.</span><span class="n">ArrayList</span><span class="p">(</span><span class="kt">i32</span><span class="p">))</span><span class="w"> </span><span class="kt">void</span><span class="w"> </span><span class="p">{</span>
|
||||
<a id="__codelineno-9-3" name="__codelineno-9-3" href="#__codelineno-9-3"></a><span class="w"> </span><span class="c1">// 当爬到第 n 阶时，方案数量加 1</span>
|
||||
<a id="__codelineno-9-4" name="__codelineno-9-4" href="#__codelineno-9-4"></a><span class="w"> </span><span class="k">if</span><span class="w"> </span><span class="p">(</span><span class="n">state</span><span class="w"> </span><span class="o">==</span><span class="w"> </span><span class="n">n</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
|
||||
<a id="__codelineno-9-5" name="__codelineno-9-5" href="#__codelineno-9-5"></a><span class="w"> </span><span class="n">res</span><span class="p">.</span><span class="n">items</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">res</span><span class="p">.</span><span class="n">items</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span><span class="w"> </span><span class="o">+</span><span class="w"> </span><span class="mi">1</span><span class="p">;</span>
|
||||
<a id="__codelineno-9-6" name="__codelineno-9-6" href="#__codelineno-9-6"></a><span class="w"> </span><span class="p">}</span>
|
||||
<a id="__codelineno-9-7" name="__codelineno-9-7" href="#__codelineno-9-7"></a><span class="w"> </span><span class="c1">// 遍历所有选择</span>
|
||||
<a id="__codelineno-9-8" name="__codelineno-9-8" href="#__codelineno-9-8"></a><span class="w"> </span><span class="k">for</span><span class="w"> </span><span class="p">(</span><span class="n">choices</span><span class="p">)</span><span class="w"> </span><span class="o">|</span><span class="n">choice</span><span class="o">|</span><span class="w"> </span><span class="p">{</span>
|
||||
<a id="__codelineno-9-9" name="__codelineno-9-9" href="#__codelineno-9-9"></a><span class="w"> </span><span class="c1">// 剪枝：不允许越过第 n 阶</span>
|
||||
<a id="__codelineno-9-10" name="__codelineno-9-10" href="#__codelineno-9-10"></a><span class="w"> </span><span class="k">if</span><span class="w"> </span><span class="p">(</span><span class="n">state</span><span class="w"> </span><span class="o">+</span><span class="w"> </span><span class="n">choice</span><span class="w"> </span><span class="o">></span><span class="w"> </span><span class="n">n</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
|
||||
<a id="__codelineno-9-11" name="__codelineno-9-11" href="#__codelineno-9-11"></a><span class="w"> </span><span class="k">break</span><span class="p">;</span>
|
||||
<a id="__codelineno-9-12" name="__codelineno-9-12" href="#__codelineno-9-12"></a><span class="w"> </span><span class="p">}</span>
|
||||
<a id="__codelineno-9-13" name="__codelineno-9-13" href="#__codelineno-9-13"></a><span class="w"> </span><span class="c1">// 尝试：做出选择，更新状态</span>
|
||||
<a id="__codelineno-9-14" name="__codelineno-9-14" href="#__codelineno-9-14"></a><span class="w"> </span><span class="n">backtrack</span><span class="p">(</span><span class="n">choices</span><span class="p">,</span><span class="w"> </span><span class="n">state</span><span class="w"> </span><span class="o">+</span><span class="w"> </span><span class="n">choice</span><span class="p">,</span><span class="w"> </span><span class="n">n</span><span class="p">,</span><span class="w"> </span><span class="n">res</span><span class="p">);</span>
|
||||
<a id="__codelineno-9-15" name="__codelineno-9-15" href="#__codelineno-9-15"></a><span class="w"> </span><span class="c1">// 回退</span>
|
||||
<a id="__codelineno-9-16" name="__codelineno-9-16" href="#__codelineno-9-16"></a><span class="w"> </span><span class="p">}</span>
|
||||
<a id="__codelineno-9-17" name="__codelineno-9-17" href="#__codelineno-9-17"></a><span class="p">}</span>
|
||||
<a id="__codelineno-9-18" name="__codelineno-9-18" href="#__codelineno-9-18"></a>
|
||||
<a id="__codelineno-9-19" name="__codelineno-9-19" href="#__codelineno-9-19"></a><span class="p">[</span><span class="n">class</span><span class="p">]{}</span><span class="o">-</span><span class="p">[</span><span class="n">func</span><span class="p">]{</span><span class="n">climbingStairsBacktrack</span><span class="p">}</span>
|
||||
</code></pre></div>
|
||||
</div>
|
||||
<div class="tabbed-block">
|
||||
@ -2441,7 +2538,7 @@
|
||||
</div>
|
||||
</div>
|
||||
</div>
|
||||
<h2 id="1311">13.1.1. 方法一：暴力搜索<a class="headerlink" href="#1311" title="Permanent link">¶</a></h2>
|
||||
<h2 id="1411">14.1.1. 方法一：暴力搜索<a class="headerlink" href="#1411" title="Permanent link">¶</a></h2>
|
||||
<p>回溯算法通常并不显式地对问题进行拆解，而是将问题看作一系列决策步骤，通过试探和剪枝，搜索所有可能的解。</p>
|
||||
<p>对于本题，我们可以尝试将问题拆解为更小的子问题。设爬到第 <span class="arithmatex">\(i\)</span> 阶共有 <span class="arithmatex">\(dp[i]\)</span> 种方案，那么 <span class="arithmatex">\(dp[i]\)</span> 就是原问题，其子问题包括:</p>
|
||||
<div class="arithmatex">\[
|
||||
@ -2557,9 +2654,18 @@ dp[i] = dp[i-1] + dp[i-2]
|
||||
</code></pre></div>
|
||||
</div>
|
||||
<div class="tabbed-block">
|
||||
<div class="highlight"><span class="filename">climbing_stairs_dfs.zig</span><pre><span></span><code><a id="__codelineno-20-1" name="__codelineno-20-1" href="#__codelineno-20-1"></a><span class="p">[</span><span class="n">class</span><span class="p">]{}</span><span class="o">-</span><span class="p">[</span><span class="n">func</span><span class="p">]{</span><span class="n">dfs</span><span class="p">}</span>
|
||||
<a id="__codelineno-20-2" name="__codelineno-20-2" href="#__codelineno-20-2"></a>
|
||||
<a id="__codelineno-20-3" name="__codelineno-20-3" href="#__codelineno-20-3"></a><span class="p">[</span><span class="n">class</span><span class="p">]{}</span><span class="o">-</span><span class="p">[</span><span class="n">func</span><span class="p">]{</span><span class="n">climbingStairsDFS</span><span class="p">}</span>
|
||||
<div class="highlight"><span class="filename">climbing_stairs_dfs.zig</span><pre><span></span><code><a id="__codelineno-20-1" name="__codelineno-20-1" href="#__codelineno-20-1"></a><span class="c1">// 搜索</span>
|
||||
<a id="__codelineno-20-2" name="__codelineno-20-2" href="#__codelineno-20-2"></a><span class="k">fn</span><span class="w"> </span><span class="n">dfs</span><span class="p">(</span><span class="n">i</span><span class="o">:</span><span class="w"> </span><span class="kt">usize</span><span class="p">)</span><span class="w"> </span><span class="kt">i32</span><span class="w"> </span><span class="p">{</span>
|
||||
<a id="__codelineno-20-3" name="__codelineno-20-3" href="#__codelineno-20-3"></a><span class="w"> </span><span class="c1">// 已知 dp[1] 和 dp[2] ，返回之</span>
|
||||
<a id="__codelineno-20-4" name="__codelineno-20-4" href="#__codelineno-20-4"></a><span class="w"> </span><span class="k">if</span><span class="w"> </span><span class="p">(</span><span class="n">i</span><span class="w"> </span><span class="o">==</span><span class="w"> </span><span class="mi">1</span><span class="w"> </span><span class="k">or</span><span class="w"> </span><span class="n">i</span><span class="w"> </span><span class="o">==</span><span class="w"> </span><span class="mi">2</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
|
||||
<a id="__codelineno-20-5" name="__codelineno-20-5" href="#__codelineno-20-5"></a><span class="w"> </span><span class="k">return</span><span class="w"> </span><span class="nb">@intCast</span><span class="p">(</span><span class="n">i</span><span class="p">);</span>
|
||||
<a id="__codelineno-20-6" name="__codelineno-20-6" href="#__codelineno-20-6"></a><span class="w"> </span><span class="p">}</span>
|
||||
<a id="__codelineno-20-7" name="__codelineno-20-7" href="#__codelineno-20-7"></a><span class="w"> </span><span class="c1">// dp[i] = dp[i-1] + dp[i-2]</span>
|
||||
<a id="__codelineno-20-8" name="__codelineno-20-8" href="#__codelineno-20-8"></a><span class="w"> </span><span class="kr">var</span><span class="w"> </span><span class="n">count</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">dfs</span><span class="p">(</span><span class="n">i</span><span class="w"> </span><span class="o">-</span><span class="w"> </span><span class="mi">1</span><span class="p">)</span><span class="w"> </span><span class="o">+</span><span class="w"> </span><span class="n">dfs</span><span class="p">(</span><span class="n">i</span><span class="w"> </span><span class="o">-</span><span class="w"> </span><span class="mi">2</span><span class="p">);</span>
|
||||
<a id="__codelineno-20-9" name="__codelineno-20-9" href="#__codelineno-20-9"></a><span class="w"> </span><span class="k">return</span><span class="w"> </span><span class="n">count</span><span class="p">;</span>
|
||||
<a id="__codelineno-20-10" name="__codelineno-20-10" href="#__codelineno-20-10"></a><span class="p">}</span>
|
||||
<a id="__codelineno-20-11" name="__codelineno-20-11" href="#__codelineno-20-11"></a>
|
||||
<a id="__codelineno-20-12" name="__codelineno-20-12" href="#__codelineno-20-12"></a><span class="p">[</span><span class="n">class</span><span class="p">]{}</span><span class="o">-</span><span class="p">[</span><span class="n">func</span><span class="p">]{</span><span class="n">climbingStairsDFS</span><span class="p">}</span>
|
||||
</code></pre></div>
|
||||
</div>
|
||||
<div class="tabbed-block">
|
||||
@ -2575,7 +2681,7 @@ dp[i] = dp[i-1] + dp[i-2]
|
||||
<p align="center"> Fig. 爬楼梯对应递归树 </p>
|
||||
|
||||
<p>实际上，<strong>指数阶的时间复杂度是由于「重叠子问题」导致的</strong>。例如，问题 <span class="arithmatex">\(dp[9]\)</span> 被分解为子问题 <span class="arithmatex">\(dp[8]\)</span> 和 <span class="arithmatex">\(dp[7]\)</span> ，问题 <span class="arithmatex">\(dp[8]\)</span> 被分解为子问题 <span class="arithmatex">\(dp[7]\)</span> 和 <span class="arithmatex">\(dp[6]\)</span> ，两者都包含子问题 <span class="arithmatex">\(dp[7]\)</span> ，而子问题中又包含更小的重叠子问题，子子孙孙无穷尽也，绝大部分计算资源都浪费在这些重叠的问题上。</p>
|
||||
<h2 id="1312">13.1.2. 方法二：记忆化搜索<a class="headerlink" href="#1312" title="Permanent link">¶</a></h2>
|
||||
<h2 id="1412">14.1.2. 方法二：记忆化搜索<a class="headerlink" href="#1412" title="Permanent link">¶</a></h2>
|
||||
<p>为了提升算法效率，<strong>我们希望所有的重叠子问题都只被计算一次</strong>。具体来说，考虑借助一个数组 <code>mem</code> 来记录每个子问题的解，并在搜索过程中这样做：</p>
|
||||
<ul>
|
||||
<li>当首次计算 <span class="arithmatex">\(dp[i]\)</span> 时，我们将其记录至 <code>mem[i]</code> ，以便之后使用；</li>
|
||||
@ -2710,9 +2816,24 @@ dp[i] = dp[i-1] + dp[i-2]
|
||||
</code></pre></div>
|
||||
</div>
|
||||
<div class="tabbed-block">
|
||||
<div class="highlight"><span class="filename">climbing_stairs_dfs_mem.zig</span><pre><span></span><code><a id="__codelineno-31-1" name="__codelineno-31-1" href="#__codelineno-31-1"></a><span class="p">[</span><span class="n">class</span><span class="p">]{}</span><span class="o">-</span><span class="p">[</span><span class="n">func</span><span class="p">]{</span><span class="n">dfs</span><span class="p">}</span>
|
||||
<a id="__codelineno-31-2" name="__codelineno-31-2" href="#__codelineno-31-2"></a>
|
||||
<a id="__codelineno-31-3" name="__codelineno-31-3" href="#__codelineno-31-3"></a><span class="p">[</span><span class="n">class</span><span class="p">]{}</span><span class="o">-</span><span class="p">[</span><span class="n">func</span><span class="p">]{</span><span class="n">climbingStairsDFSMem</span><span class="p">}</span>
|
||||
<div class="highlight"><span class="filename">climbing_stairs_dfs_mem.zig</span><pre><span></span><code><a id="__codelineno-31-1" name="__codelineno-31-1" href="#__codelineno-31-1"></a><span class="c1">// 记忆化搜索</span>
|
||||
<a id="__codelineno-31-2" name="__codelineno-31-2" href="#__codelineno-31-2"></a><span class="k">fn</span><span class="w"> </span><span class="n">dfs</span><span class="p">(</span><span class="n">i</span><span class="o">:</span><span class="w"> </span><span class="kt">usize</span><span class="p">,</span><span class="w"> </span><span class="n">mem</span><span class="o">:</span><span class="w"> </span><span class="p">[]</span><span class="kt">i32</span><span class="p">)</span><span class="w"> </span><span class="kt">i32</span><span class="w"> </span><span class="p">{</span>
|
||||
<a id="__codelineno-31-3" name="__codelineno-31-3" href="#__codelineno-31-3"></a><span class="w"> </span><span class="c1">// 已知 dp[1] 和 dp[2] ，返回之</span>
|
||||
<a id="__codelineno-31-4" name="__codelineno-31-4" href="#__codelineno-31-4"></a><span class="w"> </span><span class="k">if</span><span class="w"> </span><span class="p">(</span><span class="n">i</span><span class="w"> </span><span class="o">==</span><span class="w"> </span><span class="mi">1</span><span class="w"> </span><span class="k">or</span><span class="w"> </span><span class="n">i</span><span class="w"> </span><span class="o">==</span><span class="w"> </span><span class="mi">2</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
|
||||
<a id="__codelineno-31-5" name="__codelineno-31-5" href="#__codelineno-31-5"></a><span class="w"> </span><span class="k">return</span><span class="w"> </span><span class="nb">@intCast</span><span class="p">(</span><span class="n">i</span><span class="p">);</span>
|
||||
<a id="__codelineno-31-6" name="__codelineno-31-6" href="#__codelineno-31-6"></a><span class="w"> </span><span class="p">}</span>
|
||||
<a id="__codelineno-31-7" name="__codelineno-31-7" href="#__codelineno-31-7"></a><span class="w"> </span><span class="c1">// 若存在记录 dp[i] ，则直接返回之</span>
|
||||
<a id="__codelineno-31-8" name="__codelineno-31-8" href="#__codelineno-31-8"></a><span class="w"> </span><span class="k">if</span><span class="w"> </span><span class="p">(</span><span class="n">mem</span><span class="p">[</span><span class="n">i</span><span class="p">]</span><span class="w"> </span><span class="o">!=</span><span class="w"> </span><span class="o">-</span><span class="mi">1</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
|
||||
<a id="__codelineno-31-9" name="__codelineno-31-9" href="#__codelineno-31-9"></a><span class="w"> </span><span class="k">return</span><span class="w"> </span><span class="n">mem</span><span class="p">[</span><span class="n">i</span><span class="p">];</span>
|
||||
<a id="__codelineno-31-10" name="__codelineno-31-10" href="#__codelineno-31-10"></a><span class="w"> </span><span class="p">}</span>
|
||||
<a id="__codelineno-31-11" name="__codelineno-31-11" href="#__codelineno-31-11"></a><span class="w"> </span><span class="c1">// dp[i] = dp[i-1] + dp[i-2]</span>
|
||||
<a id="__codelineno-31-12" name="__codelineno-31-12" href="#__codelineno-31-12"></a><span class="w"> </span><span class="kr">var</span><span class="w"> </span><span class="n">count</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">dfs</span><span class="p">(</span><span class="n">i</span><span class="w"> </span><span class="o">-</span><span class="w"> </span><span class="mi">1</span><span class="p">,</span><span class="w"> </span><span class="n">mem</span><span class="p">)</span><span class="w"> </span><span class="o">+</span><span class="w"> </span><span class="n">dfs</span><span class="p">(</span><span class="n">i</span><span class="w"> </span><span class="o">-</span><span class="w"> </span><span class="mi">2</span><span class="p">,</span><span class="w"> </span><span class="n">mem</span><span class="p">);</span>
|
||||
<a id="__codelineno-31-13" name="__codelineno-31-13" href="#__codelineno-31-13"></a><span class="w"> </span><span class="c1">// 记录 dp[i]</span>
|
||||
<a id="__codelineno-31-14" name="__codelineno-31-14" href="#__codelineno-31-14"></a><span class="w"> </span><span class="n">mem</span><span class="p">[</span><span class="n">i</span><span class="p">]</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">count</span><span class="p">;</span>
|
||||
<a id="__codelineno-31-15" name="__codelineno-31-15" href="#__codelineno-31-15"></a><span class="w"> </span><span class="k">return</span><span class="w"> </span><span class="n">count</span><span class="p">;</span>
|
||||
<a id="__codelineno-31-16" name="__codelineno-31-16" href="#__codelineno-31-16"></a><span class="p">}</span>
|
||||
<a id="__codelineno-31-17" name="__codelineno-31-17" href="#__codelineno-31-17"></a>
|
||||
<a id="__codelineno-31-18" name="__codelineno-31-18" href="#__codelineno-31-18"></a><span class="p">[</span><span class="n">class</span><span class="p">]{}</span><span class="o">-</span><span class="p">[</span><span class="n">func</span><span class="p">]{</span><span class="n">climbingStairsDFSMem</span><span class="p">}</span>
|
||||
</code></pre></div>
|
||||
</div>
|
||||
<div class="tabbed-block">
|
||||
@ -2727,7 +2848,7 @@ dp[i] = dp[i-1] + dp[i-2]
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<p><img alt="记忆化搜索对应递归树" src="../intro_to_dynamic_programming.assets/climbing_stairs_dfs_memo_tree.png" /></p>
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<p align="center"> Fig. 记忆化搜索对应递归树 </p>
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<h2 id="1313">13.1.3. 方法三：动态规划<a class="headerlink" href="#1313" title="Permanent link">¶</a></h2>
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<h2 id="1413">14.1.3. 方法三：动态规划<a class="headerlink" href="#1413" title="Permanent link">¶</a></h2>
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<p><strong>记忆化搜索是一种“从顶至底”的方法</strong>：我们从原问题（根节点）开始，递归地将较大子问题分解为较小子问题，直至解已知的最小子问题（叶节点）；最终通过回溯将子问题的解逐层收集，得到原问题的解。</p>
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<p><strong>我们也可以直接“从底至顶”进行求解</strong>，得到标准的动态规划解法：从最小子问题开始，迭代地求解较大子问题，直至得到原问题的解。</p>
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<p>由于动态规划不包含回溯过程，因此无需使用递归，而可以直接基于递推实现。我们初始化一个数组 <code>dp</code> 来存储子问题的解，从最小子问题开始，逐步求解较大子问题。在以下代码中，数组 <code>dp</code> 起到了记忆化搜索中数组 <code>mem</code> 相同的记录作用。</p>
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@ -2934,8 +3055,8 @@ dp[i] = dp[i-1] + dp[i-2]
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<p>总的看来，<strong>子问题分解是一种通用的算法思路，在分治、动态规划、回溯中各有特点</strong>：</p>
|
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<ul>
|
||||
<li>分治算法将原问题划分为几个独立的子问题，然后递归解决子问题，最后合并子问题的解得到原问题的解。例如，归并排序将长数组不断划分为两个短子数组，再将排序好的子数组合并为排序好的长数组。</li>
|
||||
<li>动态规划也是将原问题分解为多个子问题，但与分治算法的主要区别是，<strong>动态规划中的子问题往往不是相互独立的</strong>，原问题的解依赖于子问题的解，而子问题的解又依赖于更小的子问题的解。因此，动态规划通常会引入记忆化，保存已经解决的子问题的解，避免重复计算。</li>
|
||||
<li>回溯算法在尝试和回退中穷举所有可能的解，并通过剪枝避免不必要的搜索分支。原问题的解由一系列决策步骤构成，我们可以将每个决策步骤之后的剩余问题看作为一个子问题。</li>
|
||||
<li>动态规划也是将原问题分解为多个子问题，但与分治算法的主要区别是，<strong>动态规划中的子问题往往不是相互独立的</strong>，原问题的解依赖于子问题的解，而子问题的解又依赖于更小的子问题的解。</li>
|
||||
<li>回溯算法在尝试和回退中穷举所有可能的解，并通过剪枝避免不必要的搜索分支。原问题的解由一系列决策步骤构成，我们可以将每个决策步骤之前的子序列看作为一个子问题。</li>
|
||||
</ul>
|
||||
|
||||
|
||||
@ -3014,7 +3135,7 @@ dp[i] = dp[i-1] + dp[i-2]
|
||||
<nav class="md-footer__inner md-grid" aria-label="页脚" >
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||||
|
||||
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||||
<a href="../" class="md-footer__link md-footer__link--prev" aria-label="上一页: 13. &nbsp; 动态规划" rel="prev">
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<a href="../" class="md-footer__link md-footer__link--prev" aria-label="上一页: 14. &nbsp; 动态规划" rel="prev">
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||||
@ -3023,20 +3144,20 @@ dp[i] = dp[i-1] + dp[i-2]
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13. 动态规划
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||||
14. 动态规划
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||||
</div>
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||||
</div>
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||||
</a>
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||||
|
||||
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||||
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13.2. DP 问题特性（New）
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14.2. DP 问题特性（New）
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</div>
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