Ching 9690121403 feat(init project): add all existing files
add all existing files

Signed-off-by: Ching <loooching@gmail.com>
2022-02-02 19:04:18 +08:00

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<!DOCTYPE html><html lang="zh-CN"><head><meta charset="utf-8"><meta name="X-UA-Compatible" content="IE=edge"><title> leetcode-1013 · MarkDown</title><meta name="description" content="leetcode-1013 - Ching"><meta name="viewport" content="width=device-width, initial-scale=1"><link rel="short icon" href="/favicon.png"><link rel="stylesheet" href="/css/apollo.css"><link rel="stylesheet" href="https://fonts.googleapis.com/css?family=Source+Sans+Pro:400,600" type="text/css"><style><!-- hexo-inject:begin --><!-- hexo-inject:end -->mjx-container[jax="SVG"] {
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</style><!-- hexo-inject:begin --><!-- hexo-inject:end --></head><body><header><a href="/" class="logo-link"><img src="/logo.png"></a><ul class="nav nav-list"><li class="nav-list-item"><a href="/" target="_self" class="nav-list-link">ALL</a></li><li class="nav-list-item"><a href="/categories/leetcode/" target="_self" class="nav-list-link">LEETCODE</a></li><li class="nav-list-item"><a href="https://bearmiebear.blogspot.com" target="_blank" class="nav-list-link">BEAR</a></li><li class="nav-list-item"><a href="/atom.xml" target="_self" class="nav-list-link">RSS</a></li></ul></header><section class="container"><div class="post"><article class="post-block"><h1 class="post-title">leetcode-1013</h1><div class="post-meta"><div class="post-time">2020年3月29日</div></div><div class="post-content"><h3 id="1013-__u5C06_u6570_u7EC4_u5206_u6210_u548C_u76F8_u7B49_u7684_u4E09_u4E2A_u90E8_u5206"><a href="#1013-__u5C06_u6570_u7EC4_u5206_u6210_u548C_u76F8_u7B49_u7684_u4E09_u4E2A_u90E8_u5206" class="headerlink" title="1013. 将数组分成和相等的三个部分"></a>1013. 将数组分成和相等的三个部分</h3><p><a href="https://leetcode-cn.com/problems/partition-array-into-three-parts-with-equal-sum/" target="_blank" rel="noopener">题目</a></p>
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<p>因为是整数数组如果能均分成三份则数组和肯定是3的倍数。然后遍历数组逐端求和使得和为 sum(A)/3。</p>
<figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span>:</span></span><br><span class="line"> <span class="function"><span class="keyword">def</span> <span class="title">canThreePartsEqualSum</span><span class="params">(self, A)</span> -&gt; bool:</span></span><br><span class="line"> <span class="keyword">if</span> <span class="keyword">not</span> A:</span><br><span class="line"> <span class="keyword">return</span> <span class="keyword">False</span></span><br><span class="line"> sa = sum(A)</span><br><span class="line"> <span class="keyword">if</span> sa % <span class="number">3</span>:</span><br><span class="line"> <span class="keyword">return</span> <span class="keyword">False</span></span><br><span class="line"> s = sa // <span class="number">3</span></span><br><span class="line"> s1 = <span class="number">0</span></span><br><span class="line"> s2 = <span class="number">0</span></span><br><span class="line"></span><br><span class="line"></span><br><span class="line"> <span class="keyword">for</span> i <span class="keyword">in</span> range(len(A)):</span><br><span class="line"> s1 += A[i]</span><br><span class="line"> <span class="keyword">if</span> s1 == s <span class="keyword">and</span> (i+<span class="number">1</span>) &lt; len(A):</span><br><span class="line"> <span class="keyword">for</span> j <span class="keyword">in</span> range(len(A[i+<span class="number">1</span>:])):</span><br><span class="line"> s2 += A[i+<span class="number">1</span>+j]</span><br><span class="line"> <span class="keyword">if</span> s2 == s <span class="keyword">and</span> j+<span class="number">1</span> &lt; len(A[i+<span class="number">1</span>:]) <span class="keyword">and</span> sum(A[i+j+<span class="number">2</span>:])== s:</span><br><span class="line"> <span class="keyword">return</span> <span class="keyword">True</span></span><br><span class="line"> <span class="keyword">return</span> <span class="keyword">False</span></span><br><span class="line"><span class="comment">#60 ms 18.7 MB</span></span><br></pre></td></tr></table></figure>
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