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<link itemprop="mainEntityOfPage" href="http://yoursite.com/2018/09/08/判断一个点是否在多边形内的PNPloy算法/">
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<a class="post-title-link" href="/2018/09/08/判断一个点是否在多边形内的PNPloy算法/" itemprop="url">判断一个点是否在多边形内的PNPloy算法</a></h1>
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<p>前几天做阿里的一题面试题如下：</p>
<p>给定一个点（x,y），一个长度为n的点集{(x1,xy1),(x2,xy2),(x3,xy3)……(xn,xyn)}，判断该点是否在点集构成的多边形中，在的话输出yes，不在的话输出该点到多边形的最短距离。</p>
<p>首先判断该点是否在多边形中，就用到了标题的PNPloy算法。</p>
<p>PNPloy算法的基本思想为射线法。</p>
<p>射线法：</p>
<p><img src="http://ph6s4kugu.bkt.clouddn.com/ray.gif" alt="ray"></p>
<p><img src="http://ph6s4kugu.bkt.clouddn.com/ray2.png" alt="ray2"></p>
<p>如上图所示：</p>
<p>红点或黑点为目标点</p>
<p>一目标点为起点发射一条射线，观察射线与多边形的交点</p>
<p>如果目标点在多边形内，交点数为奇数个</p>
<p>如果目标点不在多边形内，交点数为偶数个</p>
<p>此外还有几种特殊情况：</p>
<p>1.目标点在多边形边上</p>
<p><img src="http://ph6s4kugu.bkt.clouddn.com/ray3.png" alt="1536482831840"></p>
<p>2.目标点与多边形的点重合</p>
<p><img src="http://ph6s4kugu.bkt.clouddn.com/ray4.png" alt="1536482870928"></p>
<p>3.目标点发出的射线经过多边形的点</p>
<p><img src="http://ph6s4kugu.bkt.clouddn.com/ray5.png" alt="41536482898659"></p>
<p>4.目标点发出的射线与多边形的边重合</p>
<p><img src="http://ph6s4kugu.bkt.clouddn.com/ray6.png" alt="1536483024647"></p>
<p>对这三种情况的特殊处理：</p>
<ul>
<li><p>对于情况1，需要计算目标点与两个多边形顶点的连线斜率是否相等，但是在同一直线上不代表目标点就在边上，也可能是在边的延长线上，所以还需要考虑两个多边形顶点和目标点的坐标大小关系。需要满足x1 =< x =< x2或y1<= y <= y2。</p>
</li>
<li><p>对于情况2，只需要遍历多边形的点集，判断是否重合即可</p>
</li>
<li><p>对于情况3，可以将射线与点相交看作射线与边相交。具体做法是，做一个规定，如果一个点与目标点的射线相交，那么我们便认为这个点在射线的上方。</p>
</li>
</ul>
<p><img src="http://ph6s4kugu.bkt.clouddn.com/ray7.png" alt="1536485853818"></p>
<p>如图所示：</p>
<p>对于x点，认定D点在x发出的射线上方，故射线与边AD、CD相交，交点数为偶数，所以x在多边形外</p>
<p>同理，对于y点，CD与射线不相交，BC与射线相交，交点数为奇数，所欲y在多边形内</p>
<p>对于z点，AB、BC均不与射线相交，故z在多边形外。</p>
<ul>
<li>最后对于情况4，我们在处理情况3是也顺带把情况4解决了</li>
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<h1 id="快速排序"><a href="#快速排序" class="headerlink" title="快速排序"></a>快速排序</h1><p>比较经典的一种快速排序实现</p>
<figure class="highlight plain"><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><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br></pre></td><td class="code"><pre><span class="line">public class quiksort {</span><br><span class="line"> public static void main(String args[]){</span><br><span class="line"> quiksort qs = new quiksort();</span><br><span class="line"> int[] list = new int[]{1};</span><br><span class="line"> qs.quicksort(list, 0, list.length - 1 );</span><br><span class="line"> for (int i = 0; i < list.length; i++){</span><br><span class="line"> System.out.print(list[i] + " ");</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> private void quicksort(int[] list, int i, int j){</span><br><span class="line"> if (i < j){</span><br><span class="line"> int cu = partion(list, i, j);</span><br><span class="line"> quicksort(list, i, cu -1 );</span><br><span class="line"> quicksort(list, cu +1, j);</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"></span><br><span class="line"> private int partion(int[] list, int i, int j){</span><br><span class="line"> int x = list[j];</span><br><span class="line"> int cu = i - 1;</span><br><span class="line"> for (int tmp = i; tmp < j; tmp ++){</span><br><span class="line"> if (list[tmp] < x){</span><br><span class="line"> cu++;</span><br><span class="line"> swap(list, cu, tmp);</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> cu++;</span><br><span class="line"> swap(list, cu, j);</span><br><span class="line"> return cu;</span><br><span class="line"> }</span><br><span class="line"></span><br><span class="line"> private void swap(int[] list, int i, int j){</span><br><span class="line"> int tmp = list[i];</span><br><span class="line"> list[i] = list[j];</span><br><span class="line"> list[j] = tmp;</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
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<a class="post-title-link" href="/2018/08/12/二叉树的非递归遍历/" itemprop="url">二叉树的非递归遍历</a></h1>
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<h2 id="前序遍历"><a href="#前序遍历" class="headerlink" title="前序遍历"></a>前序遍历</h2><p>没什么好说的，用一个栈暂存节点，先中后左最后右</p>
<figure class="highlight java"><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></pre></td><td class="code"><pre><span class="line"><span class="keyword">public</span> <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">public</span> <span class="keyword">static</span> <span class="keyword">void</span> <span class="title">preOrderNonRecursive</span><span class="params">(TreeNode root)</span> </span>{</span><br><span class="line"> TreeNode cur = root;</span><br><span class="line"> Stack<TreeNode> stack = <span class="keyword">new</span> Stack<>();</span><br><span class="line"> <span class="keyword">while</span> (cur != <span class="keyword">null</span> || !stack.empty()){</span><br><span class="line"> <span class="keyword">if</span> (cur != <span class="keyword">null</span>) {</span><br><span class="line"> System.out.println(cur.val);</span><br><span class="line"> stack.push(cur);</span><br><span class="line"> cur = cur.left;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">else</span>{</span><br><span class="line"> cur = stack.pop().right;</span><br><span class="line"> }</span><br><span class="line"></span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<h2 id="中序遍历"><a href="#中序遍历" class="headerlink" title="中序遍历"></a>中序遍历</h2><p>中序和前续区别不大，唯一的不同点在于先访问父节点还是先访问左子节点</p>
<figure class="highlight java"><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></pre></td><td class="code"><pre><span class="line"><span class="keyword">public</span> <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">public</span> <span class="keyword">static</span> <span class="keyword">void</span> <span class="title">OrderNonRecursive</span><span class="params">(TreeNode root)</span> </span>{</span><br><span class="line"> TreeNode cur = root;</span><br><span class="line"> Stack<TreeNode> stack = <span class="keyword">new</span> Stack<>();</span><br><span class="line"> <span class="keyword">while</span> (cur != <span class="keyword">null</span> || !stack.empty()){</span><br><span class="line"> <span class="keyword">if</span> (cur != <span class="keyword">null</span>) {</span><br><span class="line"> stack.push(cur);</span><br><span class="line"> cur = cur.left;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">else</span>{</span><br><span class="line"> System.out.println(stack.peek().val);</span><br><span class="line"> cur = stack.pop().right;</span><br><span class="line"> }</span><br><span class="line"></span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<h1 id="后续遍历"><a href="#后续遍历" class="headerlink" title="后续遍历"></a>后续遍历</h1><p>总体思路：</p>
<p>同样是用栈来进行遍历线路的控制，但是后续遍历的关键之处，也就是与上述两种遍历的不同点在于：前序遍历和中序遍历中右子节点都是晚于父节点遍历的。</p>
<p>所以前两种遍历，在经过左子节点出栈后，其父节点可以紧接着出栈，而后续遍历在左子节点出栈后，第一次访问父节点时候不出栈，需要使其右子节点入栈，等到第二次访问父节点时，也就是其右子节点出栈后才出栈。</p>
<p>所以关键就在于父节点的出栈条件。稍加思考可以发现，父节点出栈前，其右子节点先出栈，故我们可以做一个判断：但遍历到某个节点A时，如果这时候上一个处理的节点B是A的右子节点，说明这时候是第二次访问A，故A出栈。</p>
<figure class="highlight java"><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><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">public</span> <span class="class"><span class="keyword">class</span> <span class="title">Solution</span> </span>{</span><br><span class="line"></span><br><span class="line">}</span><br><span class="line"><span class="keyword">public</span> <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">public</span> <span class="keyword">static</span> <span class="keyword">void</span> <span class="title">OrderNonRecursive</span><span class="params">(TreeNode root)</span> </span>{</span><br><span class="line"> TreeNode cur = root;</span><br><span class="line"> Stack<TreeNode> stack = <span class="keyword">new</span> Stack<>();</span><br><span class="line"> <span class="keyword">while</span> (cur != <span class="keyword">null</span> || !stack.empty()){</span><br><span class="line"> <span class="keyword">if</span> (cur != <span class="keyword">null</span>) {</span><br><span class="line"> stack.push(cur);</span><br><span class="line"> cur = cur.left;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">else</span>{</span><br><span class="line"> <span class="keyword">while</span> (cur == stack.peek().right) {</span><br><span class="line"> cur = stack.pop();</span><br><span class="line"> System.out.println(cur.val);</span><br><span class="line"> <span class="keyword">if</span> (stack.empty()) </span><br><span class="line"> <span class="keyword">return</span>;</span><br><span class="line"> }</span><br><span class="line"> cur = stack.peek().right;</span><br><span class="line"> }</span><br><span class="line"></span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
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<a class="post-title-link" href="/2018/06/21/剑指offer题号38. 字符串的排列/" itemprop="url">剑指offer题号39.数组中出现次数超过一半的数字</a></h1>
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<p>数组中有一个数字出现的次数超过数组长度的一半，请找出这个数字。例如输入一个长度为9的数组{1,2,3,2,2,2,5,4,2}。由于数字2在数组中出现了5次，超过数组长度的一半，因此输出2。如果不存在则输出0。 </p>
<figure class="highlight java"><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><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">public</span> <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">public</span> <span class="keyword">static</span> <span class="keyword">void</span> <span class="title">main</span><span class="params">()</span></span>{</span><br><span class="line"> Solution solution = <span class="keyword">new</span> Solution();</span><br><span class="line"> }</span><br><span class="line"></span><br><span class="line"> <span class="keyword">private</span> ArrayList<ArrayList<Character>> ret;</span><br><span class="line"></span><br><span class="line"> <span class="keyword">public</span> ArrayList<ArrayList<Character>> permutation(String str){</span><br><span class="line"> <span class="keyword">if</span> (str.length() == <span class="number">0</span>) {</span><br><span class="line"> <span class="keyword">return</span> <span class="keyword">null</span>;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">char</span>[] chars = str.toCharArray();</span><br><span class="line"> Arrays.sort(chars);</span><br><span class="line"> backtracking(chars, <span class="keyword">new</span> <span class="keyword">boolean</span>[chars.length], <span class="keyword">new</span> ArrayList<Character>());</span><br><span class="line"> <span class="keyword">return</span> ret;</span><br><span class="line"> }</span><br><span class="line"> <span class="function"><span class="keyword">public</span> <span class="keyword">void</span> <span class="title">backtracking</span><span class="params">(<span class="keyword">char</span>[] chars, <span class="keyword">boolean</span>[] hased, ArrayList<Character> tmp)</span></span>{</span><br><span class="line"> <span class="keyword">if</span> (tmp.size() == chars.length){</span><br><span class="line"> ret.add(tmp);</span><br><span class="line"> <span class="keyword">return</span>;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">0</span> ; i < chars.length; i++){</span><br><span class="line"> <span class="keyword">if</span> (hased[i]) <span class="keyword">continue</span>;</span><br><span class="line"> <span class="keyword">if</span> (i != <span class="number">0</span> && chars[i] == chars[i-<span class="number">1</span>] && !hased[i-<span class="number">1</span>]) <span class="keyword">continue</span>; <span class="comment">//此处是最为关键的一步，对于出现重复的情况，比如aaa，只要用到第二个a而没有用到第一个a时，说明此时已经时重复了，故不做处理。</span></span><br><span class="line"> hased[i] = <span class="keyword">true</span>;</span><br><span class="line"> tmp.add(chars[i]);</span><br><span class="line"> backtracking(chars, hased, tmp);</span><br><span class="line"> tmp.remove(tmp.size());</span><br><span class="line"> hased[i] = <span class="keyword">false</span>;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>简单粗暴的方法，方法的时间复杂度为o（n！）</p>
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<a class="post-title-link" href="/2018/06/19/docker学习之发布自己的python项目镜像/" itemprop="url">docker学习之发布自己的python项目镜像</a></h1>
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<h2 id="1-环境配置"><a href="#1-环境配置" class="headerlink" title="1.环境配置"></a>1.环境配置</h2><p>安装centos</p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">$ docker pull centos:latest</span><br></pre></td></tr></table></figure>
<p>更新yum</p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">$ yum -y update</span><br></pre></td></tr></table></figure>
<p>-y: 提前在系统确认(y/n)?中选择yes</p>
<p>安装yum-utils，一个yum扩展集合</p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">$ yum -y install yum-utils</span><br></pre></td></tr></table></figure>
<p>安装development，允许你通过源码生成和编译软件</p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">$ yum -y groupinstall development</span><br></pre></td></tr></table></figure>
<p><strong><em>安装python3</em></strong></p>
<p>安装IUS，提供一些比较新的RPM包</p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">$ yum -y install https://centos7.iuscommunity.org/ius-release.rpm</span><br></pre></td></tr></table></figure>
<p>安装python</p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">$ yum -y install python36u</span><br></pre></td></tr></table></figure>
<p>检查pythpn是否安装成功</p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">$ python3.6 -V</span><br></pre></td></tr></table></figure>
<p>安装pip</p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">$ yum -y install python36u-pip</span><br></pre></td></tr></table></figure>
<p>最后安装python36u-devel</p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">$ yum -y install python36u-devel</span><br></pre></td></tr></table></figure>
<h2 id="安装ssh"><a href="#安装ssh" class="headerlink" title="安装ssh"></a>安装ssh</h2><figure class="highlight plain"><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></pre></td><td class="code"><pre><span class="line">$ yum install initscripts</span><br><span class="line">$ yum install openssh*</span><br><span class="line">$ systemctl enable sshd</span><br><span class="line">$ systemctl start sshd</span><br><span class="line">$ service sshd start</span><br></pre></td></tr></table></figure>
<p>查看ssh服务是否运行</p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">$ yum install net-tools</span><br><span class="line">$ netstat -ntlp</span><br></pre></td></tr></table></figure>
<p>结果如下，说明成功运行</p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">Proto Recv-Q Send-Q Local Address Foreign Address State PID/Program name</span><br><span class="line">tcp 0 0 0.0.0.0:22 0.0.0.0:* LISTEN 12742/sshd</span><br></pre></td></tr></table></figure>
<h2 id="保存镜像"><a href="#保存镜像" class="headerlink" title="保存镜像"></a>保存镜像</h2><p>退出容器</p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">$ exit</span><br></pre></td></tr></table></figure>
<p>保存镜像</p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">docker commit -m "install openssh" -a "57ing" 1101a03a13c2 centos:ssh</span><br></pre></td></tr></table></figure>
<p>-m：本次提交介绍（install openssh）</p>
<p>-a：本次提交用户信息（57ing）</p>
<p>1101a03a13c2： 容器id</p>
<p>centos:ssh： 镜像名</p>
<h2 id="启动新的容器并打通22端口"><a href="#启动新的容器并打通22端口" class="headerlink" title="启动新的容器并打通22端口"></a>启动新的容器并打通22端口</h2><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">docker run -d -p 50001:22 centos:ssh /usr/sbin/sshd -D</span><br></pre></td></tr></table></figure>
<h2 id="可通过ssh连接进入容器内部进行各种操作"><a href="#可通过ssh连接进入容器内部进行各种操作" class="headerlink" title="可通过ssh连接进入容器内部进行各种操作"></a>可通过ssh连接进入容器内部进行各种操作</h2><h2 id="建立项目后一键启动项目"><a href="#建立项目后一键启动项目" class="headerlink" title="建立项目后一键启动项目"></a>建立项目后一键启动项目</h2><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">$ docker run -it -p 57411:5000 centos:ssh python3.6 /root/flask/main.py</span><br></pre></td></tr></table></figure>
<h2 id="将建立的镜像导出和导入"><a href="#将建立的镜像导出和导入" class="headerlink" title="将建立的镜像导出和导入"></a>将建立的镜像导出和导入</h2><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">$ docker save -o E:\docker.tar centos:ssh</span><br><span class="line">$ docker load < E:\docker.tar</span><br></pre></td></tr></table></figure>
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