<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[hello2sys_cs的专栏]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/hello2sys_cs</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; hello2sys_cs]]></copyright><item><title><![CDATA[《天才在左疯子在右》]]></title><link>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8801216</link><guid>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8801216</guid><author>hello2sys_cs</author><pubDate>Sun, 14 Apr 2013 21:05:48 +0800</pubDate><description><![CDATA[这一天都是昏昏沉沉的，没有精神，晚上重新看了一下这本书，第一次看还是在遥远的大二，那时候觉得非常非常好看，看完后还和某个大神同学讨论，跟他讲里面的案例。其实心理学真的很奇妙，有很多你想不到的有趣案例，精神病人的世界未必就是恐怖怪诞的，有些或许会很温情，就像这本书里面讲的《永远永远》故事，再读还是觉得非常感动，可能因为我爷爷的缘故，他也有几乎差不多一样的经历，不过不是跟我奶奶了，是另外一个人。天才和]]></description><category></category></item><item><title><![CDATA[我看《送你一颗子弹》]]></title><link>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8684812</link><guid>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8684812</guid><author>hello2sys_cs</author><pubDate>Sun, 17 Mar 2013 21:19:57 +0800</pubDate><description><![CDATA[这本书断断续续花了一个多星期给看完了，刚开始觉得刘瑜写的东西很琐屑，读到后面发现其实文字很诙谐，她应该是个很有幽默感的人，有幽默感的人的生活态度和知识面应该不差，再说她也是个大牛的人。政治学博士，第三性别，呵呵其实从她的文字里面，我看到一个小女人，一个理性的、睿智的、幽默大方的、渴望爱的小女人，或许还有很多特质我无法用匮乏的成语来形容。话说回来，她喜欢看电影，而我却不那么热衷花几个小时看电影而宁愿]]></description><category></category></item><item><title><![CDATA[判断数组中的数值是否连续]]></title><link>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8681027</link><guid>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8681027</guid><author>hello2sys_cs</author><pubDate>Sat, 16 Mar 2013 13:52:51 +0800</pubDate><description><![CDATA[#include
#include

bool IsContinuous1(int *arr,int N)
{/*整数数列，元素取值为0...65535中任意一个不重复的数，0可以出现多次
	1、元素值可以乱序。
	2、0可以匹配任何数值。
	3、0可以多次出现。
	4、全0算连续、只有一个非0也算连续。
 */
	int min = -1,max = -1;

	for(int i = 0;i]]></description><category></category></item><item><title><![CDATA[编程之美-从无头单链表中删除结点]]></title><link>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8680863</link><guid>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8680863</guid><author>hello2sys_cs</author><pubDate>Sat, 16 Mar 2013 12:51:03 +0800</pubDate><description><![CDATA[void DeleteRandomNode(node *pCurrent)
{//从无头单链表中删除结点

	assert(pCurrent != NULL);
	node *pNext = pCurrent->next;
	if(pNext != NULL)
	{
		pCurrent->next = pNext->next;
		pCurrent->data = pNext->data;]]></description><category></category></item><item><title><![CDATA[编程之美-字符串移位包含]]></title><link>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8680784</link><guid>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8680784</guid><author>hello2sys_cs</author><pubDate>Sat, 16 Mar 2013 12:14:11 +0800</pubDate><description><![CDATA[int StrInStr1(char *src,char *des)
{//方法一：穷举src循环移位的字符串组合，每个都跟des比较

	int len = strlen(src);
	for( int i = 0; i < len; ++i)
	{
		char tempchar = src[0];
		for( int j = 0; j < len - 1;++j)     //循环左移]]></description><category></category></item><item><title><![CDATA[我读《ZOO》、《夏天烟花我的尸体》]]></title><link>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8679831</link><guid>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8679831</guid><author>hello2sys_cs</author><pubDate>Fri, 15 Mar 2013 22:53:26 +0800</pubDate><description><![CDATA[乙一的两本知名怪诞小说《zoo》、《夏天烟花我的尸体》都拜读过了，初衷是抱着看恐怖小说的心情去看的，看完了才发现其实不恐怖，用怪诞来形容很贴切。
       在卓越上买书的时候一开始是贪便宜买的两本书，都打5折才十几块钱就200多页了，看完才发现虽然期望没有我想的那么高，但是还是有阅读的享受的，算是感觉了另外一种文风吧。看网上的书评是挺好的，说了一大堆的好话、一大堆的有N多专业术语的读后感，到]]></description><category></category></item><item><title><![CDATA[找出数组中唯一的重复元素]]></title><link>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8678162</link><guid>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8678162</guid><author>hello2sys_cs</author><pubDate>Fri, 15 Mar 2013 15:20:20 +0800</pubDate><description><![CDATA[数组a[N]，1至N-1这N-1个数存放在a[N]中，其中某个数重复一次。找出这个数void xor_findDup1(int *a,int N)
{//方法一：求和
	int tmp1 = 0;
	int tmp2 = 0;

	for(int i = 0;i < N - 1; ++i)
	{//tmp1为1...N-1的和,tmp2为数组所有元素的和

		tmp1 += (i + 1);]]></description><category></category></item><item><title><![CDATA[编程之美-最大公约数]]></title><link>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8677601</link><guid>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8677601</guid><author>hello2sys_cs</author><pubDate>Fri, 15 Mar 2013 13:44:34 +0800</pubDate><description><![CDATA[int gcd1(int x,int y)
{//欧几里得辗转相除法

	return (!y)?x:gcd1(y,x % y);
}


int gcd2(int x,int y)
{//能除的尽x,y的数，也能除的尽x-y,y
	if( x < y)
		return gcd2(y,x);
	if( y == 0 )
		return x;
	else
		return gcd2(x - y,]]></description><category></category></item><item><title><![CDATA[判断一个较大的单链表是否有环及求环的入口点]]></title><link>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8673356</link><guid>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8673356</guid><author>hello2sys_cs</author><pubDate>Thu, 14 Mar 2013 15:13:58 +0800</pubDate><description><![CDATA[/*判断单链表中是否有环*/

struct ListType
{
	int data;
	struct ListType *next;
};


typedef struct ListType *list;

int IsLoop(list s)
{//双指针法

	list fast = s;
	list slow = s;
	
	if(fast == NULL)
		return -1;]]></description><category></category></item><item><title><![CDATA[编程之美-数组循环移位]]></title><link>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8672520</link><guid>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8672520</guid><author>hello2sys_cs</author><pubDate>Thu, 14 Mar 2013 12:39:01 +0800</pubDate><description><![CDATA[#include

void RightShift1(int *arr,int N,int K)
{//循环右移

	K %= N;
	while(K--)
	{
		int t = arr[N - 1];
		for(int i = N - 1; i > 0;--i)
		{
			arr[i] = arr[i - 1];
		}
		arr[0] = t;
	}
}

void Reverse]]></description><category></category></item><item><title><![CDATA[编程之美-寻找发帖水王]]></title><link>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8667937</link><guid>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8667937</guid><author>hello2sys_cs</author><pubDate>Wed, 13 Mar 2013 13:10:18 +0800</pubDate><description><![CDATA[Type Find(Type *ID,int N)
{//寻找发帖"水王",假设只有一个
	Type candidate;
	int nTimes,i;

	for( i = nTimes = 0; i < N; i++)
	{
		if(nTimes == 0)
		{
			candidate = ID[i];
			nTimes = 1;
		}
		else
		{
			if(candi]]></description><category></category></item><item><title><![CDATA[编程之美—不要被阶乘吓倒]]></title><link>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8667804</link><guid>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8667804</guid><author>hello2sys_cs</author><pubDate>Wed, 13 Mar 2013 12:25:19 +0800</pubDate><description><![CDATA[int FactorialZeroNum1(int N)
{//N!末尾有多少个0，求1..N各个数的5的因子有多少个
	int ret = 0;
	for(int i = 1 ; i <= N ; ++i)
	{
		int j = i;
		while( j % 5 == 0)
		{
			ret++;
			j /= 5;
		}
	}

	return ret;
}

int Facto]]></description><category></category></item><item><title><![CDATA[sizeof(&b)与VC++中的bug]]></title><link>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8664236</link><guid>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8664236</guid><author>hello2sys_cs</author><pubDate>Tue, 12 Mar 2013 16:05:40 +0800</pubDate><description><![CDATA[int main()
{//2013/3/12
//sizeof(&b)在VC++6.0中为400，而在Devc++中则是4，&b的类型是int (*)[100]，明显是一个指针，应该是4  
	int b[100];
	int *c = &b;  //编译错误 cannot convert `int (*)[100]' to `int*' in initialization 
	int (*a)]]></description><category></category></item><item><title><![CDATA[在O(n)时间复杂度内找出数组中的众数]]></title><link>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8663822</link><guid>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8663822</guid><author>hello2sys_cs</author><pubDate>Tue, 12 Mar 2013 15:10:32 +0800</pubDate><description><![CDATA[bool FindMostApperse(int *num,int len,int &candidate)
{
	int count = 0;
	for(int i = 0;i < len;++i)
	{
		if(count == 0)                         //目前得到的数出现次数为0时，更换另外一个数
		{
			candidate = num[i];
			co]]></description><category></category></item><item><title><![CDATA[找出数组中重复次数最多的数]]></title><link>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8663624</link><guid>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8663624</guid><author>hello2sys_cs</author><pubDate>Tue, 12 Mar 2013 14:40:57 +0800</pubDate><description><![CDATA[#include
#include
#include

using namespace std;

//以空间换时间
int GetMaxNum(int *arr,int len,int &num)
{//找出数组arr中最大的数及其下标

	int index = arr[0];
	int i;
	for( i = 0;i <= len;++i)             //注意这里是闭区间，理]]></description><category></category></item><item><title><![CDATA[编程珠玑第十四章—堆]]></title><link>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8662716</link><guid>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8662716</guid><author>hello2sys_cs</author><pubDate>Tue, 12 Mar 2013 11:29:42 +0800</pubDate><description><![CDATA[习题：
5、将箱序列组织成堆，父结点代表了它的子节点中最不满的结点的箱的剩余空间，堆顶应该是最大容量的箱。以”首次适应“启发式方法按序考虑权值，在决定往哪里放新权值时候，尽量往左搜索，只要左边最不满的箱子有足够的空间放该权值就尽量往左，迫不得已才往右。当把权值放入箱子后，箱子容量减少，向下调整使其保持堆性质。这样所需时间正比于O(log(n))。
6、让结点i指向i+1和2i结点。就像组织成一]]></description><category></category></item><item><title><![CDATA[判断两个单链表（无环）是否交叉]]></title><link>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8660810</link><guid>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8660810</guid><author>hello2sys_cs</author><pubDate>Mon, 11 Mar 2013 20:32:45 +0800</pubDate><description><![CDATA[bool IsIntersect(Node *list1,Node *list2,Node *&value)
{//判断两个单链表(无环)是否交叉
	value = NULL;
	if(NULL == list1 || NULL == list2)
		return false;

	Node *temp = list1,*temp2 = list2;
	int len1 = 0,len2 = 0]]></description><category></category></item><item><title><![CDATA[删除单链表中的重复结点]]></title><link>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8660732</link><guid>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8660732</guid><author>hello2sys_cs</author><pubDate>Mon, 11 Mar 2013 20:14:25 +0800</pubDate><description><![CDATA[方法一：递归

link delSame(link head)
{/*	不变式：list{k..n-1}没有重复的元素，加入list{k-1}后，为了保持不变式
	要从list{k-1..n-1}两两比较，如果有相同元素必定在开头两个*/

	link pointer,temp = head;
	if(head->next == NULL)
		return head;
	head->next =]]></description><category></category></item><item><title><![CDATA[合并两个有序链表（非交叉）]]></title><link>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8660642</link><guid>https://blog.youkuaiyun.com/hello2sys_cs/article/details/8660642</guid><author>hello2sys_cs</author><pubDate>Mon, 11 Mar 2013 19:48:15 +0800</pubDate><description><![CDATA[Node *MergeRecursive(Node *head1,Node *head2)
{//递归合并两个有序链表

	if(head1 == NULL)
		return head2;
	if(head2 == NULL)
		return head1;
	Node *head = NULL;
	if(head1->data data)
	{
		head = head1;
		head->]]></description><category></category></item></channel></rss>