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10.2 &nbsp; 二分查找插入点
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10.3 &nbsp; 二分查找边界
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第 12 章 &nbsp; 分治
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12.1 &nbsp; 分治算法
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12.2 &nbsp; 分治搜索策略
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12.3 &nbsp; 构建树问题
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12.4 &nbsp; 汉诺塔问题
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12.5 &nbsp; 小结
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第 14 章 &nbsp; 动态规划
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14.1 &nbsp; 初探动态规划
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14.1 &nbsp; 初探动态规划
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14.2 &nbsp; DP 问题特性
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14.3 &nbsp; DP 解题思路
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14.4 &nbsp; 0-1 背包问题
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14.5 &nbsp; 完全背包问题
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14.6 &nbsp; 编辑距离问题
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14.7 &nbsp; 小结
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第 15 章 &nbsp; 贪心
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15.1 &nbsp; 贪心算法
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15.2 &nbsp; 分数背包问题
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15.3 &nbsp; 最大容量问题
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15.4 &nbsp; 最大切分乘积问题
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15.5 &nbsp; 小结
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<p>给定一个共有 <span class="arithmatex">\(n\)</span> 阶的楼梯,你每步可以上 <span class="arithmatex">\(1\)</span> 阶或者 <span class="arithmatex">\(2\)</span> 阶,请问有多少种方案可以爬到楼顶。</p>
</div>
<p>如图 14-1 所示,对于一个 <span class="arithmatex">\(3\)</span> 阶楼梯,共有 <span class="arithmatex">\(3\)</span> 种方案可以爬到楼顶。</p>
<p><img alt="爬到第 3 阶的方案数量" src="../intro_to_dynamic_programming.assets/climbing_stairs_example.png" /></p>
<p><a class="glightbox" href="../intro_to_dynamic_programming.assets/climbing_stairs_example.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="爬到第 3 阶的方案数量" src="../intro_to_dynamic_programming.assets/climbing_stairs_example.png" /></a></p>
<p align="center"> 图 14-1 &nbsp; 爬到第 3 阶的方案数量 </p>
<p>本题的目标是求解方案数量,<strong>我们可以考虑通过回溯来穷举所有可能性</strong>。具体来说,将爬楼梯想象为一个多轮选择的过程:从地面出发,每轮选择上 <span class="arithmatex">\(1\)</span> 阶或 <span class="arithmatex">\(2\)</span> 阶,每当到达楼梯顶部时就将方案数量加 <span class="arithmatex">\(1\)</span> ,当越过楼梯顶部时就将其剪枝。</p>
@ -3852,7 +3679,7 @@ dp[i-1], dp[i-2], \dots, dp[2], dp[1]
dp[i] = dp[i-1] + dp[i-2]
\]</div>
<p>这意味着在爬楼梯问题中,各个子问题之间存在递推关系,<strong>原问题的解可以由子问题的解构建得来</strong>。图 14-2 展示了该递推关系。</p>
<p><img alt="方案数量递推关系" src="../intro_to_dynamic_programming.assets/climbing_stairs_state_transfer.png" /></p>
<p><a class="glightbox" href="../intro_to_dynamic_programming.assets/climbing_stairs_state_transfer.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="方案数量递推关系" src="../intro_to_dynamic_programming.assets/climbing_stairs_state_transfer.png" /></a></p>
<p align="center"> 图 14-2 &nbsp; 方案数量递推关系 </p>
<p>我们可以根据递推公式得到暴力搜索解法。以 <span class="arithmatex">\(dp[n]\)</span> 为起始点,<strong>递归地将一个较大问题拆解为两个较小问题的和</strong>,直至到达最小子问题 <span class="arithmatex">\(dp[1]\)</span><span class="arithmatex">\(dp[2]\)</span> 时返回。其中,最小子问题的解是已知的,即 <span class="arithmatex">\(dp[1] = 1\)</span><span class="arithmatex">\(dp[2] = 2\)</span> ,表示爬到第 <span class="arithmatex">\(1\)</span><span class="arithmatex">\(2\)</span> 阶分别有 <span class="arithmatex">\(1\)</span><span class="arithmatex">\(2\)</span> 种方案。</p>
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<p>图 14-3 展示了暴力搜索形成的递归树。对于问题 <span class="arithmatex">\(dp[n]\)</span> ,其递归树的深度为 <span class="arithmatex">\(n\)</span> ,时间复杂度为 <span class="arithmatex">\(O(2^n)\)</span> 。指数阶属于爆炸式增长,如果我们输入一个比较大的 <span class="arithmatex">\(n\)</span> ,则会陷入漫长的等待之中。</p>
<p><img alt="爬楼梯对应递归树" src="../intro_to_dynamic_programming.assets/climbing_stairs_dfs_tree.png" /></p>
<p><a class="glightbox" href="../intro_to_dynamic_programming.assets/climbing_stairs_dfs_tree.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="爬楼梯对应递归树" src="../intro_to_dynamic_programming.assets/climbing_stairs_dfs_tree.png" /></a></p>
<p align="center"> 图 14-3 &nbsp; 爬楼梯对应递归树 </p>
<p>观察图 14-3 <strong>指数阶的时间复杂度是由于“重叠子问题”导致的</strong>。例如 <span class="arithmatex">\(dp[9]\)</span> 被分解为 <span class="arithmatex">\(dp[8]\)</span><span class="arithmatex">\(dp[7]\)</span> <span class="arithmatex">\(dp[8]\)</span> 被分解为 <span class="arithmatex">\(dp[7]\)</span><span class="arithmatex">\(dp[6]\)</span> ,两者都包含子问题 <span class="arithmatex">\(dp[7]\)</span></p>
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<p>观察图 14-4 <strong>经过记忆化处理后,所有重叠子问题都只需被计算一次,时间复杂度被优化至 <span class="arithmatex">\(O(n)\)</span></strong> ,这是一个巨大的飞跃。</p>
<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>
<p><a class="glightbox" href="../intro_to_dynamic_programming.assets/climbing_stairs_dfs_memo_tree.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="记忆化搜索对应递归树" src="../intro_to_dynamic_programming.assets/climbing_stairs_dfs_memo_tree.png" /></a></p>
<p align="center"> 图 14-4 &nbsp; 记忆化搜索对应递归树 </p>
<h2 id="1413">14.1.3 &nbsp; 方法三:动态规划<a class="headerlink" href="#1413" title="Permanent link">&para;</a></h2>
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<p>图 14-5 模拟了以上代码的执行过程。</p>
<p><img alt="爬楼梯的动态规划过程" src="../intro_to_dynamic_programming.assets/climbing_stairs_dp.png" /></p>
<p><a class="glightbox" href="../intro_to_dynamic_programming.assets/climbing_stairs_dp.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="爬楼梯的动态规划过程" src="../intro_to_dynamic_programming.assets/climbing_stairs_dp.png" /></a></p>
<p align="center"> 图 14-5 &nbsp; 爬楼梯的动态规划过程 </p>
<p>与回溯算法一样,动态规划也使用“状态”概念来表示问题求解的某个特定阶段,每个状态都对应一个子问题以及相应的局部最优解。例如,爬楼梯问题的状态定义为当前所在楼梯阶数 <span class="arithmatex">\(i\)</span></p>
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