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第 6 章 &nbsp; 哈希表
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第 8 章 &nbsp;
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第 9 章 &nbsp;
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9.3 &nbsp; 图的遍历
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第 10 章 &nbsp; 搜索
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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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10.4 &nbsp; 哈希优化策略
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10.5 &nbsp; 重识搜索算法
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第 11 章 &nbsp; 排序
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11.2 &nbsp; 选择排序
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11.3 &nbsp; 冒泡排序
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11.4 &nbsp; 插入排序
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11.5 &nbsp; 快速排序
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11.6 &nbsp; 归并排序
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11.7 &nbsp; 堆排序
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11.8 &nbsp; 桶排序
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11.9 &nbsp; 计数排序
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11.10 &nbsp; 基数排序
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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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第 13 章 &nbsp; 回溯
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第 14 章 &nbsp; 动态规划
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第 15 章 &nbsp; 贪心
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<h1 id="147">14.7 &nbsp; 小结<a class="headerlink" href="#147" title="Permanent link">&para;</a></h1>
<ul>
<li>动态规划对问题进行分解,并通过存储子问题的解来规避重复计算,提高 计算效率。</li>
<li>不考虑时间的前提下,所有动态规划问题都可以用回溯(暴力搜索)进行求解,但递归树中存在大量的重叠子问题,效率极低。通过引入记忆化列表,可以存储所有计算过的子问题的解,从而保证重叠子问题只被计算一次。</li>
<li>记忆化递归是一种从顶至底的递归式解法,而与之对应的动态规划是一种从底至顶的递推式解法,其如同“填写表格”一样。由于当前状态仅依赖某些局部状态,因此我们可以消除 <span class="arithmatex">\(dp\)</span> 表的一个维度,从而降低空间复杂度。</li>
<li>子问题分解是一种通用的算法思路,在分治、动态规划、回溯中具有不同的性质。</li>
<li>动态规划问题有三大特性:重叠子问题、最优子结构、无后效性。</li>
<li>如果原问题的最优解可以从子问题的最优解构建得来,则它就具有最优子结构。</li>
<li>无后效性指对于一个状态,其未来发展只与该状态有关,而与过去经历的所有状态无关。许多组合优化问题不具有无后效性,无法使用动态规划快速求解。</li>
</ul>
<p><strong>背包问题</strong></p>
<ul>
<li>背包问题是最典型的动态规划问题之一,具有 0-1 背包、完全背包、多重背包等变种。</li>
<li>0-1 背包的状态定义为前 <span class="arithmatex">\(i\)</span> 个物品在剩余容量为 <span class="arithmatex">\(c\)</span> 的背包中的最大价值。根据不放入背包和放入背包两种决策,可得到最优子结构,并构建出状态转移方程。在空间优化中,由于每个状态依赖正上方和左上方的状态,因此需要倒序遍历列表,避免左上方状态被覆盖。</li>
<li>完全背包问题的每种物品的选取数量无限制,因此选择放入物品的状态转移与 0-1 背包问题不同。由于状态依赖正上方和正左方的状态,因此在空间优化中应当正序遍历。</li>
<li>零钱兑换问题是完全背包问题的一个变种。它从求“最大”价值变为求“最小”硬币数量,因此状态转移方程中的 <span class="arithmatex">\(\max()\)</span> 应改为 <span class="arithmatex">\(\min()\)</span> 。从追求“不超过”背包容量到追求“恰好”凑出目标金额,因此使用 <span class="arithmatex">\(amt + 1\)</span> 来表示“无法凑出目标金额”的无效解。</li>
<li>零钱兑换 II 问题从求“最少硬币数量”改为求“硬币组合数量”,状态转移方程相应地从 <span class="arithmatex">\(\min()\)</span> 改为求和运算符。</li>
</ul>
<p><strong>编辑距离问题</strong></p>
<ul>
<li>编辑距离Levenshtein 距离)用于衡量两个字符串之间的相似度,其定义为从一个字符串到另一个字符串的最少编辑步数,编辑操作包括添加、删除、替换。</li>
<li>编辑距离问题的状态定义为将 <span class="arithmatex">\(s\)</span> 的前 <span class="arithmatex">\(i\)</span> 个字符更改为 <span class="arithmatex">\(t\)</span> 的前 <span class="arithmatex">\(j\)</span> 个字符所需的最少编辑步数。当 <span class="arithmatex">\(s[i] \ne t[j]\)</span> 时,具有三种决策:添加、删除、替换,它们都有相应的剩余子问题。据此便可以找出最优子结构与构建状态转移方程。而当 <span class="arithmatex">\(s[i] = t[j]\)</span> 时,无须编辑当前字符。</li>
<li>在编辑距离中,状态依赖其正上方、正左方、左上方的状态,因此空间优化后正序或倒序遍历都无法正确地进行状态转移。为此,我们利用一个变量暂存左上方状态,从而转化到与完全背包问题等价的情况,可以在空间优化后进行正序遍历。</li>
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