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lcp/10/index.html

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<!-- solution:start -->
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<h3 id="_3">方法一</h3>
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<div class="tabbed-set tabbed-alternate" data-tabs="1:5"><input checked="checked" id="__tabbed_1_1" name="__tabbed_1" type="radio" /><input id="__tabbed_1_2" name="__tabbed_1" type="radio" /><input id="__tabbed_1_3" name="__tabbed_1" type="radio" /><input id="__tabbed_1_4" name="__tabbed_1" type="radio" /><input id="__tabbed_1_5" name="__tabbed_1" type="radio" /><div class="tabbed-labels"><label for="__tabbed_1_1">Python3</label><label for="__tabbed_1_2">Java</label><label for="__tabbed_1_3">C++</label><label for="__tabbed_1_4">Go</label><label for="__tabbed_1_5">TypeScript</label></div>
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<div class="tabbed-set tabbed-alternate" data-tabs="1:6"><input checked="checked" id="__tabbed_1_1" name="__tabbed_1" type="radio" /><input id="__tabbed_1_2" name="__tabbed_1" type="radio" /><input id="__tabbed_1_3" name="__tabbed_1" type="radio" /><input id="__tabbed_1_4" name="__tabbed_1" type="radio" /><input id="__tabbed_1_5" name="__tabbed_1" type="radio" /><input id="__tabbed_1_6" name="__tabbed_1" type="radio" /><div class="tabbed-labels"><label for="__tabbed_1_1">Python3</label><label for="__tabbed_1_2">Java</label><label for="__tabbed_1_3">C++</label><label for="__tabbed_1_4">Go</label><label for="__tabbed_1_5">TypeScript</label><label for="__tabbed_1_6">Swift</label></div>
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<div class="tabbed-content">
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<div class="highlight"><table class="highlighttable"><tr><td class="linenos"><div class="linenodiv"><pre><span></span><span class="normal"> 1</span>
@@ -85089,6 +85089,72 @@ <h3 id="_3">方法一</h3>
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<span class="p">}</span>
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</code></pre></div></td></tr></table></div>
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</div>
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<div class="tabbed-block">
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<div class="highlight"><table class="highlighttable"><tr><td class="linenos"><div class="linenodiv"><pre><span></span><span class="normal"> 1</span>
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<span class="normal">32</span></pre></div></td><td class="code"><div><pre><span></span><code><span class="cm">/**</span>
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<span class="cm">* public class TreeNode {</span>
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<span class="cm">* public var val: Int</span>
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<span class="cm">* public var left: TreeNode?</span>
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<span class="cm">* public var right: TreeNode?</span>
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<span class="cm">* public init() { self.val = 0; self.left = nil; self.right = nil; }</span>
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<span class="cm">* public init(_ val: Int) { self.val = val; self.left = nil; self.right = nil; }</span>
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<span class="cm">* public init(_ val: Int, _ left: TreeNode?, _ right: TreeNode?) {</span>
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<span class="cm">* self.val = val</span>
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<span class="cm">* self.left = left</span>
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<span class="cm">* self.right = right</span>
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<span class="cm">* }</span>
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<span class="cm">* }</span>
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<span class="cm">*/</span>
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<span class="kd">class</span> <span class="nc">Solution</span> <span class="p">{</span>
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<span class="kd">func</span> <span class="nf">minimalExecTime</span><span class="p">(</span><span class="kc">_</span> <span class="n">root</span><span class="p">:</span> <span class="n">TreeNode</span><span class="p">?)</span> <span class="p">-&gt;</span> <span class="nb">Double</span> <span class="p">{</span>
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<span class="k">return</span> <span class="n">dfs</span><span class="p">(</span><span class="n">root</span><span class="p">)[</span><span class="mi">1</span><span class="p">]</span>
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<span class="p">}</span>
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<span class="kd">private</span> <span class="kd">func</span> <span class="nf">dfs</span><span class="p">(</span><span class="kc">_</span> <span class="n">root</span><span class="p">:</span> <span class="n">TreeNode</span><span class="p">?)</span> <span class="p">-&gt;</span> <span class="p">[</span><span class="nb">Double</span><span class="p">]</span> <span class="p">{</span>
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<span class="k">guard</span> <span class="kd">let</span> <span class="nv">root</span> <span class="p">=</span> <span class="n">root</span> <span class="k">else</span> <span class="p">{</span> <span class="k">return</span> <span class="p">[</span><span class="mf">0.0</span><span class="p">,</span> <span class="mf">0.0</span><span class="p">]</span> <span class="p">}</span>
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<span class="kd">let</span> <span class="nv">left</span> <span class="p">=</span> <span class="n">dfs</span><span class="p">(</span><span class="n">root</span><span class="p">.</span><span class="kr">left</span><span class="p">)</span>
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<span class="kd">let</span> <span class="nv">right</span> <span class="p">=</span> <span class="n">dfs</span><span class="p">(</span><span class="n">root</span><span class="p">.</span><span class="kr">right</span><span class="p">)</span>
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<span class="kd">let</span> <span class="nv">sum</span> <span class="p">=</span> <span class="kr">left</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span> <span class="o">+</span> <span class="kr">right</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span> <span class="o">+</span> <span class="nb">Double</span><span class="p">(</span><span class="n">root</span><span class="p">.</span><span class="n">val</span><span class="p">)</span>
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<span class="kd">let</span> <span class="nv">time</span> <span class="p">=</span> <span class="bp">max</span><span class="p">(</span><span class="bp">max</span><span class="p">(</span><span class="kr">left</span><span class="p">[</span><span class="mi">1</span><span class="p">],</span> <span class="kr">right</span><span class="p">[</span><span class="mi">1</span><span class="p">]),</span> <span class="p">(</span><span class="kr">left</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span> <span class="o">+</span> <span class="kr">right</span><span class="p">[</span><span class="mi">0</span><span class="p">])</span> <span class="o">/</span> <span class="mi">2</span><span class="p">)</span> <span class="o">+</span> <span class="nb">Double</span><span class="p">(</span><span class="n">root</span><span class="p">.</span><span class="n">val</span><span class="p">)</span>
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<span class="k">return</span> <span class="p">[</span><span class="n">sum</span><span class="p">,</span> <span class="n">time</span><span class="p">]</span>
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<span class="p">}</span>
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<span class="p">}</span>
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</code></pre></div></td></tr></table></div>
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</div>
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<nav>
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<a href="https://github.com/klever34" class="md-author" title="@klever34">
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<img src="https://avatars.githubusercontent.com/u/12745225?v=4&size=72" alt="klever34">
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</a>
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<a href="https://github.com/yanglbme" class="md-author" title="@yanglbme">
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<img src="https://avatars.githubusercontent.com/u/21008209?v=4&size=72" alt="yanglbme">

lcp/11/index.html

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@@ -84853,7 +84853,7 @@ <h3 id="_3">方法一:哈希表</h3>
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<p>如果有 $n$ 个人的能力值相同,每个人有 $n$ 种不同的位置,那么每个人在原位的概率是 $\frac{1}{n}$,那么合起来的期望就是 $1$。</p>
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<p>因此,我们只需要统计不同的能力值的个数,即为答案。</p>
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<p>时间复杂度 $O(n)$,空间复杂度 $O(n)$。其中 $n$ 为数组 <code>scores</code> 的长度。</p>
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<div class="tabbed-set tabbed-alternate" data-tabs="1:5"><input checked="checked" id="__tabbed_1_1" name="__tabbed_1" type="radio" /><input id="__tabbed_1_2" name="__tabbed_1" type="radio" /><input id="__tabbed_1_3" name="__tabbed_1" type="radio" /><input id="__tabbed_1_4" name="__tabbed_1" type="radio" /><input id="__tabbed_1_5" name="__tabbed_1" type="radio" /><div class="tabbed-labels"><label for="__tabbed_1_1">Python3</label><label for="__tabbed_1_2">Java</label><label for="__tabbed_1_3">C++</label><label for="__tabbed_1_4">Go</label><label for="__tabbed_1_5">TypeScript</label></div>
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<div class="tabbed-set tabbed-alternate" data-tabs="1:6"><input checked="checked" id="__tabbed_1_1" name="__tabbed_1" type="radio" /><input id="__tabbed_1_2" name="__tabbed_1" type="radio" /><input id="__tabbed_1_3" name="__tabbed_1" type="radio" /><input id="__tabbed_1_4" name="__tabbed_1" type="radio" /><input id="__tabbed_1_5" name="__tabbed_1" type="radio" /><input id="__tabbed_1_6" name="__tabbed_1" type="radio" /><div class="tabbed-labels"><label for="__tabbed_1_1">Python3</label><label for="__tabbed_1_2">Java</label><label for="__tabbed_1_3">C++</label><label for="__tabbed_1_4">Go</label><label for="__tabbed_1_5">TypeScript</label><label for="__tabbed_1_6">Swift</label></div>
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<span class="p">}</span>
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</code></pre></div></td></tr></table></div>
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</div>
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<span class="kd">func</span> <span class="nf">expectNumber</span><span class="p">(</span><span class="kc">_</span> <span class="n">scores</span><span class="p">:</span> <span class="p">[</span><span class="nb">Int</span><span class="p">])</span> <span class="p">-&gt;</span> <span class="nb">Int</span> <span class="p">{</span>
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<span class="kd">let</span> <span class="nv">uniqueScores</span> <span class="p">=</span> <span class="n">Set</span><span class="p">(</span><span class="n">scores</span><span class="p">)</span>
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<span class="k">return</span> <span class="n">uniqueScores</span><span class="p">.</span><span class="bp">count</span>
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<span class="p">}</span>
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<span class="p">}</span>
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</code></pre></div></td></tr></table></div>
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</div>
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<a href="https://github.com/klever34" class="md-author" title="@klever34">
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<img src="https://avatars.githubusercontent.com/u/12745225?v=4&size=72" alt="klever34">
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lcp/6/index.html

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<p>我们可以发现,每堆力扣币拿完的最少次数,等于该堆力扣币数量除以 $2$ 向上取整的结果之和。</p>
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<p>因此,我们只需要遍历每堆力扣币 $x_i$,计算每堆力扣币拿完的最少次数 $\left \lceil x_i/2 \right \rceil$,然后累加即可。</p>
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<p>时间复杂度 $O(n)$,其中 $n$ 是数组 $coins$ 的长度。空间复杂度 $O(1)$。</p>
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<span class="kd">func</span> <span class="nf">minCount</span><span class="p">(</span><span class="kc">_</span> <span class="n">coins</span><span class="p">:</span> <span class="p">[</span><span class="nb">Int</span><span class="p">])</span> <span class="p">-&gt;</span> <span class="nb">Int</span> <span class="p">{</span>
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<span class="kd">var</span> <span class="nv">ans</span> <span class="p">=</span> <span class="mi">0</span>
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<span class="k">for</span> <span class="n">x</span> <span class="k">in</span> <span class="n">coins</span> <span class="p">{</span>
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<span class="n">ans</span> <span class="o">+=</span> <span class="p">(</span><span class="n">x</span> <span class="o">+</span> <span class="mi">1</span><span class="p">)</span> <span class="o">&gt;&gt;</span> <span class="mi">1</span>
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</div>
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<!-- solution:end -->
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<img src="https://avatars.githubusercontent.com/u/21008209?v=4&size=72" alt="yanglbme">
@@ -85024,12 +85049,13 @@ <h3 id="_3">方法一:数学</h3>
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<a href="https://github.com/doocs/leetcode/blob/main/lcp/LCP%2006.%20%E6%8B%BF%E7%A1%AC%E5%B8%81/README.md" class="md-author md-author--more">
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