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	<title>递归 - 四号程序员</title>
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		<title>原地反置链表(非递归和递归)</title>
		<link>https://www.coder4.com/archives/3262</link>
					<comments>https://www.coder4.com/archives/3262#respond</comments>
		
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		<pubDate>Wed, 30 May 2012 06:50:46 +0000</pubDate>
				<category><![CDATA[算法&数据结构]]></category>
		<category><![CDATA[反置链表]]></category>
		<category><![CDATA[递归]]></category>
		<category><![CDATA[非递归]]></category>
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					<description><![CDATA[链表定义如下： struct Node { struct Node* next; int data; }; 非递归反置链表如下： struct Node* ll_reverse(struct Node* head) { if(head==NULL) { return NULL; } else { struct Node* pre = NULL; struct Node* next = NULL; struct Node* ptr =[......] 继续阅读]]></description>
		
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		<title>数据结构重读 – 递归与汉诺塔</title>
		<link>https://www.coder4.com/archives/3105</link>
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		<dc:creator><![CDATA[coder4]]></dc:creator>
		<pubDate>Fri, 04 May 2012 15:03:34 +0000</pubDate>
				<category><![CDATA[算法&数据结构]]></category>
		<category><![CDATA[数据结构]]></category>
		<category><![CDATA[栈]]></category>
		<category><![CDATA[汉诺塔]]></category>
		<category><![CDATA[递归]]></category>
		<category><![CDATA[重读]]></category>
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					<description><![CDATA[堆栈与递归是相辅相成的。 比如Fibnacci数列就是递归定义： Fib(n) = Fib(n-1) + Fib(n-2) (n&#62;=2) Fib(0) = 0 Fib(1) = 1 Fibnacci数列：1, 1, 2, 3, 5, 8, 13, 21, .... 说到这里再写个Fibnacci的通项公公式： 这个还是很无敌的…… 然后再一个例子是书上的Ackerman函数： &#160; 转回经典的汉诺塔问题：假设有三个命名为X，Y[......] 继续阅读]]></description>
		
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		<title>《数据结构》读书笔记 第三章 非循环顺序队列2</title>
		<link>https://www.coder4.com/archives/224</link>
					<comments>https://www.coder4.com/archives/224#respond</comments>
		
		<dc:creator><![CDATA[coder4]]></dc:creator>
		<pubDate>Sat, 26 Dec 2009 06:03:57 +0000</pubDate>
				<category><![CDATA[算法&数据结构]]></category>
		<category><![CDATA[递归]]></category>
		<category><![CDATA[非循环顺序队列]]></category>
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					<description><![CDATA[比起第一个顺序非循环队列，这个是浪费空间，节约时间版本～ Queue.h:定义了队列的基本操作 #include &#60;iostream&#62; enum {OK=0,WRONG=-1}; enum {QIS=10,QI=2}; typedef int Elem; typedef int Status; typedef struct { &#160;Elem *base; &#160;int front; &#160;int rear; &#160;int qsize; }Q[......] 继续阅读]]></description>
		
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		<title>迷宫问题的递归c++实现</title>
		<link>https://www.coder4.com/archives/220</link>
					<comments>https://www.coder4.com/archives/220#respond</comments>
		
		<dc:creator><![CDATA[coder4]]></dc:creator>
		<pubDate>Sat, 26 Dec 2009 06:02:00 +0000</pubDate>
				<category><![CDATA[算法&数据结构]]></category>
		<category><![CDATA[迷宫]]></category>
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					<description><![CDATA[迷宫问题问题描述：把一只小白鼠放入一个迷宫，问能否找到一条路让小白鼠出来。迷宫用数组表示。能通过则给出图&#8230;&#8230; 不同于上次用栈实现，这里的&#8220;栈&#8221;自动地由系统分配，So&#8230;&#8230; &#160; #include &#60;iostream&#62; struct Pos { &#160;int x; &#160;int y; }; int map[7][7]={{0,0,0,0,0,0,0},{0,1,0,1,1,1,0}[......] 继续阅读]]></description>
		
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		<title>《数据结构》递归实现汉诺塔</title>
		<link>https://www.coder4.com/archives/216</link>
					<comments>https://www.coder4.com/archives/216#respond</comments>
		
		<dc:creator><![CDATA[coder4]]></dc:creator>
		<pubDate>Sat, 26 Dec 2009 05:59:29 +0000</pubDate>
				<category><![CDATA[算法&数据结构]]></category>
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					<description><![CDATA[main.cpp：汉诺塔实现。 思路分析： 若n=1，则只需把盘子从a挪动到c 若n&#62;1， （1）把前n-1个盘子从a经过c挪动到b （2）把第n个盘子从a挪动到c （3）把前n-1个盘子从b经过a挪动回c 实现： #include &#60;fstream&#62; #include &#60;iostream&#62; using namespace std; ofstream fout(&#34;out.txt&#34;); void Move(int n,char x,char y[......] 继续阅读]]></description>
		
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		<title>整数划分 递归</title>
		<link>https://www.coder4.com/archives/201</link>
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		<dc:creator><![CDATA[coder4]]></dc:creator>
		<pubDate>Sat, 26 Dec 2009 05:51:08 +0000</pubDate>
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					<description><![CDATA[#include &#60;iostream&#62; char *result=(char *)malloc(sizeof(char)*200); int fun(int n,int m); int main() { &#160;std::cout&#60;&#160;return 0; } int fun(int n,int m) { &#160;if(n&#60;0&#124;&#124;m&#60;0) return 0; &#160;if(n==1&#124;&#124;m==1) return 1; &#038;nbsp[......] 继续阅读]]></description>
		
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		<title>链表的就地反转 c语言实现</title>
		<link>https://www.coder4.com/archives/112</link>
					<comments>https://www.coder4.com/archives/112#respond</comments>
		
		<dc:creator><![CDATA[coder4]]></dc:creator>
		<pubDate>Thu, 26 Nov 2009 08:53:01 +0000</pubDate>
				<category><![CDATA[算法&数据结构]]></category>
		<category><![CDATA[反转]]></category>
		<category><![CDATA[递归]]></category>
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					<description><![CDATA[#include typedef struct node { int num; struct node *next; }*LINK,NODE; LINK link_rev(LINK head); LINK link_cr(int num); void link_show(LINK head); int main() { LINK *p,head,head1; head=link_cr(5); printf("Now create 5 NODE LINK,print:");[......] 继续阅读]]></description>
		
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		<title>俄式乘法 c语言实现</title>
		<link>https://www.coder4.com/archives/105</link>
					<comments>https://www.coder4.com/archives/105#respond</comments>
		
		<dc:creator><![CDATA[coder4]]></dc:creator>
		<pubDate>Thu, 26 Nov 2009 08:43:22 +0000</pubDate>
				<category><![CDATA[算法&数据结构]]></category>
		<category><![CDATA[c语言]]></category>
		<category><![CDATA[俄式乘法]]></category>
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					<description><![CDATA[#include long muti(long n,long m); int main() { printf("%ld",muti(1000,1000)); return 0; } long muti(long n,long m) { if(n==1) return m; else { if(n%2==0) { return muti(n>>1,m1,m1,m[......] 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>
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		<category><![CDATA[递归]]></category>
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					<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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		<title>全排列的C语言描述-递归实现</title>
		<link>https://www.coder4.com/archives/84</link>
					<comments>https://www.coder4.com/archives/84#respond</comments>
		
		<dc:creator><![CDATA[coder4]]></dc:creator>
		<pubDate>Thu, 26 Nov 2009 08:30:49 +0000</pubDate>
				<category><![CDATA[算法&数据结构]]></category>
		<category><![CDATA[c语言]]></category>
		<category><![CDATA[全排列]]></category>
		<category><![CDATA[实现]]></category>
		<category><![CDATA[递归]]></category>
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					<description><![CDATA[递归实现全排列: #include #define N 4 int used[N]; int out[N]; void f(int *num,int n,int p); int main() { int num[N],i; for(i=0;i[......] 继续阅读]]></description>
		
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