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	<title>算法 - 四号程序员</title>
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		<title>[转]A Tour of Machine Learning Algorithms 机器学习算法回顾</title>
		<link>https://www.coder4.com/archives/4527</link>
					<comments>https://www.coder4.com/archives/4527#respond</comments>
		
		<dc:creator><![CDATA[coder4]]></dc:creator>
		<pubDate>Fri, 20 Mar 2015 15:21:35 +0000</pubDate>
				<category><![CDATA[机器学习]]></category>
		<category><![CDATA[算法]]></category>
		<guid isPermaLink="false">http://www.coder4.com/?p=4527</guid>

					<description><![CDATA[转载自：A Tour of Machine Learning Algorithms After we understand the type of machine learning problem we are working with, we can think about the type of data to collect and the types of machine learning algorithms we can try. In this post we take a to[......] 继续阅读]]></description>
		
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		<title>求单向链表倒数第 k 个结点</title>
		<link>https://www.coder4.com/archives/3295</link>
					<comments>https://www.coder4.com/archives/3295#respond</comments>
		
		<dc:creator><![CDATA[coder4]]></dc:creator>
		<pubDate>Sat, 02 Jun 2012 04:05:55 +0000</pubDate>
				<category><![CDATA[算法&数据结构]]></category>
		<category><![CDATA[求单向链表倒数第 k 个结点]]></category>
		<category><![CDATA[算法]]></category>
		<category><![CDATA[面试]]></category>
		<guid isPermaLink="false">http://www.coder4.com/?p=3295</guid>

					<description><![CDATA[题目：输入一个单向链表，输出该链表中倒数第 k 个结点。 传统做法是获取到链表的长度n，然后走n-k布，此方法弱爆了…… 做法： 1、指定一个新的指针，走k步。如果k布之内就到了NULL，显然无倒数第k这说法，返回错误即可。 2、新指针到位后，和root一起next，知道新指针到了NULL。此时输出root指针当前的data即可。 int ll_last_k(struct Node* root, int k) { // Make two ptr diff k di[......] 继续阅读]]></description>
		
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		<title>找唯一重复的元素</title>
		<link>https://www.coder4.com/archives/2950</link>
					<comments>https://www.coder4.com/archives/2950#comments</comments>
		
		<dc:creator><![CDATA[coder4]]></dc:creator>
		<pubDate>Sun, 26 Feb 2012 12:54:24 +0000</pubDate>
				<category><![CDATA[算法&数据结构]]></category>
		<category><![CDATA[算法]]></category>
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					<description><![CDATA[1-1000放在含有1001个元素的数组中，只有唯一的一个元素值重复，其它均只出现一次。每个数组元素只能访问一次，设计一个算法，将它找出来；不用辅助存储空间，能否设计一个算法实现？ 普通青年解法： 1+2+....+1000-&#62;5050 a1+a2+....+a1001-&#62;X x - 5050 -&#62; n(重复) 文艺青年解法： a1^a2^.....a1001 = n（对所有的元素求一遍异或，就是最终重复的元素） 证明也不复杂，见：[......] 继续阅读]]></description>
		
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		<title>关于Random Shuffling算法。</title>
		<link>https://www.coder4.com/archives/2948</link>
					<comments>https://www.coder4.com/archives/2948#comments</comments>
		
		<dc:creator><![CDATA[coder4]]></dc:creator>
		<pubDate>Sun, 26 Feb 2012 09:56:05 +0000</pubDate>
				<category><![CDATA[C && C++]]></category>
		<category><![CDATA[算法&数据结构]]></category>
		<category><![CDATA[Random Shuffling]]></category>
		<category><![CDATA[洗牌]]></category>
		<category><![CDATA[算法]]></category>
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					<description><![CDATA[其实就是随机洗牌。 Knuth给过一个算法，为代码如下： 注意：随机数不是1~n，而是i~n！！ For i = 1 to n Pick a random integer j from i to n Swap A[i] and A[j] 关于为什么如此，吾等码农就不了解了，等大神来证明吧……[......] 继续阅读]]></description>
		
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		<title>大量数据取k个最大值并排序</title>
		<link>https://www.coder4.com/archives/2920</link>
					<comments>https://www.coder4.com/archives/2920#comments</comments>
		
		<dc:creator><![CDATA[coder4]]></dc:creator>
		<pubDate>Tue, 14 Feb 2012 07:47:45 +0000</pubDate>
				<category><![CDATA[C && C++]]></category>
		<category><![CDATA[算法&数据结构]]></category>
		<category><![CDATA[k最大]]></category>
		<category><![CDATA[算法]]></category>
		<category><![CDATA[面试]]></category>
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					<description><![CDATA[需求是这样的，我们都知道，在信息检索中，经常要取top-k(一共k，而不是第k)个得分最大的文档，并且从大到小排序。 而且文档规模很大，最少也要上千万。 话说这是一道很可以拿来面试的题啊。 我们不考虑Hadoop神马的，就说说单机怎么搞。 最傻的做法就是把1000万个都存储下来，然后sort，然后取min(k, vec.size())。 这样有两个缺点： 1、内存占用非常大，其实我们只要保留最大的1000个，但这样就要保存N个。在1000万的测试中，它要占用68M[......] 继续阅读]]></description>
		
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		<item>
		<title>程序员编程艺术系列</title>
		<link>https://www.coder4.com/archives/2154</link>
					<comments>https://www.coder4.com/archives/2154#respond</comments>
		
		<dc:creator><![CDATA[coder4]]></dc:creator>
		<pubDate>Sat, 17 Sep 2011 08:11:49 +0000</pubDate>
				<category><![CDATA[算法&数据结构]]></category>
		<category><![CDATA[算法]]></category>
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					<description><![CDATA[推荐一下，很强大…… http://blog.csdn.net/v_JULY_v/article/category/784066[......] 继续阅读]]></description>
		
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			<slash:comments>0</slash:comments>
		
		
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		<title>Bloom Filter实现的一些文章</title>
		<link>https://www.coder4.com/archives/2083</link>
					<comments>https://www.coder4.com/archives/2083#respond</comments>
		
		<dc:creator><![CDATA[coder4]]></dc:creator>
		<pubDate>Wed, 17 Aug 2011 15:04:21 +0000</pubDate>
				<category><![CDATA[算法&数据结构]]></category>
		<category><![CDATA[BKDR]]></category>
		<category><![CDATA[Bloom Filter]]></category>
		<category><![CDATA[ELF]]></category>
		<category><![CDATA[实现]]></category>
		<category><![CDATA[算法]]></category>
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					<description><![CDATA[1、给出了Java实现，用Random做为一致性哈希算法。。。 http://blog.locut.us/2008/01/12/a-decent-stand-alone-java-bloom-filter-implementation/ 2、分析比较到位： http://blog.csdn.net/jiaomeng/article/details/1495500 3、这个写的也不错 http://www.cnblogs.com/heaad/archive/2011/01/02/[......] 继续阅读]]></description>
		
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		<item>
		<title>Miller-Rabin算法</title>
		<link>https://www.coder4.com/archives/570</link>
					<comments>https://www.coder4.com/archives/570#comments</comments>
		
		<dc:creator><![CDATA[coder4]]></dc:creator>
		<pubDate>Sat, 20 Mar 2010 07:29:47 +0000</pubDate>
				<category><![CDATA[算法&数据结构]]></category>
		<category><![CDATA[Miller]]></category>
		<category><![CDATA[Rabin]]></category>
		<category><![CDATA[概率算法]]></category>
		<category><![CDATA[算法]]></category>
		<category><![CDATA[素数判断]]></category>
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					<description><![CDATA[&#160; Input: n &#62; 3, an odd integer to be tested for primality; Input: k, a parameter that determines the accuracy of the test Output: composite if n is composite, otherwise probably prime write n &#8722; 1 as 2s&#183;d with d odd by factoring p[......] 继续阅读]]></description>
		
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		<title>棋盘覆盖问题</title>
		<link>https://www.coder4.com/archives/101</link>
					<comments>https://www.coder4.com/archives/101#respond</comments>
		
		<dc:creator><![CDATA[coder4]]></dc:creator>
		<pubDate>Thu, 26 Nov 2009 08:42:20 +0000</pubDate>
				<category><![CDATA[算法&数据结构]]></category>
		<category><![CDATA[c语言]]></category>
		<category><![CDATA[棋盘覆盖]]></category>
		<category><![CDATA[算法]]></category>
		<category><![CDATA[递归]]></category>
		<guid isPermaLink="false">http://www.coder4.com/?p=101</guid>

					<description><![CDATA[/* 标准文档模板 */ #include "Stdio.h" #include "Conio.h" #define N 16 void Chess(int tr,int tc,int dr,int dc,int size); void Show(); int board[N+1][N+1]; int main() { /* 此处添加你自己的代码 */ board[1][1]=-1; Chess(1,1,1,1,N); Show(); getch(); retu[......] 继续阅读]]></description>
		
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