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从n个items中随机选择m个的问题

simplyzhao — Mon, 18 Jul 2011 01:32:00 GMT

问题来源: �~�程珠玑

解法一:
遍历�q�n个items�Q��y妙地利用概率来筛�?

void
generate_random_m_from_n(int n, int m)
{
    int i, remaining, select = m;
    srand(time(NULL));
    for(i=0; i<n; i++) {
        remaining = n - i;
        if(rand()%remaining < select) {
            printf("%d\t", i);
            --select;
        }
    }
    printf("\n");
}

解法�?
shuffle�Q�即随机�z�牌�E�序�Q�然后选择前m个items卛_��
代码参�? http://blog.fuqcool.com/2011/04/17/algorithm-shuffle.html

�z�牌��法的一�U�实�?/a>

作者：fuqcool 发布旉��Q?011-04-17 23:16:02 分类�Q?algorithms

最�q�自己在做一个小的程序，需要把一个集合里面的元素全部随机地打散。自己想了一个方法，复杂度是n�Q�觉得不太快。后来参照了一下python关于shuffle的算法，发现我的�Ҏ��跟它的是一��L��Q�连python的代码都�q�么写，可能已经没有办法再快了吧�Q?/p>

下面��来介绍�z�牌��法�Q�用C语言描述�?/p>

��法的前提是有一个��生随机数的函�?/p>

// Generates a random integer between beg and end.
int GetRandomNumber(int beg, int end);

�q�有一个交换函数�?/p>

// Swap a and b.
void Swap(int a, int b);

上面两个函数我就不写出实��C��Q�因��文章的重点在于��法的讨论�?/p>

假设我们有一堆扑克牌�Q�怎么才能把这副牌完全打�ؕ呢？计算机当然不能像人手那样�z�牌。但是它可以产生随机敎ͼ�随机��C��一副牌中抽��Z��张牌是可以的。既然这样那��好办了�Q�我们不停地从牌堆中随机抽取一张扑克牌�Q�然后把�q�些牌堆��h��Q�直到原来的牌堆只剩下一张牌的时候�ؓ止。这样不��完成了�z�牌的动作了吗�?/p>

下面是C代码�Q?/p>

int Shuffle(int[] a, int len)
{
    for (int i = len - 1; i > 0; i--)
    {
        // Select an element from index 0 to i randomly;
        int index = GetRandomNumber(0, i);
        // exchange a[i] with a[index]
        Swap(a[index], a[i]);
    }
}

��Z��也脓出python的random单元关于shuffle的实玎ͼ�

def shuffle(self, x, random=None, int=int):
    """x, random=random.random -> shuffle list x in place; return None.

    Optional arg random is a 0-argument function returning a random
    float in [0.0, 1.0); by default, the standard random.random.
    """

    if random is None:
        random = self.random
    for i in reversed(xrange(1, len(x))):
        # pick an element in x[:i+1] with which to exchange x[i]
        j = int(random() * (i+1))
        x[i], x[j] = x[j], x[i]

simplyzhao 2011-07-18 09:32 发表评论

simplyzhao — Thu, 14 Jul 2011 02:13:00 GMT

来源:
http://coolshell.cn/articles/3345.html

Software Engineer

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struct foo {

char a;

char* b;

};

simplyzhao 2011-07-14 10:13 发表评论

扑克牌的��子

simplyzhao — Tue, 12 Jul 2011 12:33:00 GMT

题目来源:
http://zhedahht.blog.163.com/blog/static/25411174200951262930831/

题目�Q�从扑克牌中随机�?/span>5张牌�Q�判断是不是一个顺子，卌��5张牌是不是连�l�的�?/span>2-10为数字本�w�，A�?/span>1�Q?/span>J�?/span>11�Q?/span>Q�?/span>12�Q?/span>K�?/span>13�Q�而大��王可以看成��L��数字�?/span>

思�\一:
我们需要把扑克牌的背景抽象成计��机语言。不难想象，我们可以�?/span>5张牌看成�?/span>5个数字组成的数组。大��王是特�D�的数字�Q�我们不妨把它们都当�?/span>0�Q�这样和其他扑克牌代表的数字��׃��重复了�?/span>

接下来我们来分析怎样判断5个数字是不是�q�箋的。最直观的是�Q�我们把数组排序。但值得注意的是�Q�由�?/span>0可以当成��L��数字�Q�我们可以用0去补满数�l�中的空�~�。也��是排序之后的数�l�不是连�l�的�Q�即盔R��的两个数字相隔若�q�个数字�Q�但如果我们有��够的0可以补满�q�两个数字的�I�缺�Q�这个数�l�实际上�q�是�q�箋的。�D个例子，数组排序之后�?/span>{0�Q?/span>1�Q?/span>3�Q?/span>4�Q?/span>5}。在1�?/span>3之间�I�缺了一�?/span>2�Q�刚好我们有一�?/span>0�Q�也��是我们可以它当�?/span>2��d��补这个空�~��?/span>

于是我们需要做三�g事情�Q�把数组排序�Q�统计数�l�中0的个敎ͼ��l�计排序之后的数�l�相��L��字之间的�I�缺��L��。如果空�~�的��L��于或者等�?/span>0的个敎ͼ�那么�q�个数组��是�q�箋的；反之则不�q�箋。最后，我们�q�需要注意的是，如果数组中的�?/span>0数字重复出现�Q�则该数�l�不是连�l�的。换成扑克牌的描�q�方式，��是如果一副牌里含有对子，则不可能是顺子�?br />
更好的思�\�?
1�Q�确�?张牌中除�?�Q�其余数字没有重复的�Q�可以用表统计的�Ҏ��Q?
2) 满��q�样的逻辑�Q�（max�Q�min分别代表5张牌中的�?以外的最大值最��|��
       如果没有0�Q�则max-min=4�Q�则为顺子，否则不是
       如果有一�?�Q�则max-min=4或�?�Q�则为顺子，否则不是
       如果有两�?�Q�则max-min=4或�?或�?�Q�则为顺子，否则不是

最大值和最��值在1�Q�中��可以获得，�q�样��׃��用排序了

simplyzhao 2011-07-12 20:33 发表评论

simplyzhao — Wed, 22 Jun 2011 11:41:00 GMT

题目来源:
http://zhedahht.blog.163.com/blog/static/25411174201102642136998/

题目�Q�六�Q�：�q�行下列C++代码�Q�输��Z��么？

struct Point3D

{

int x;

int y;

int z;

};

int _tmain(int argc, _TCHAR* argv[])

{

Point3D* pPoint = NULL;

int offset = (int)(&(pPoint)->z);

printf("%d", offset);

return 0;

}

�{�案�Q?/span>输出8。由于在pPoint->z的前面加上了取地址�W�号�Q�运行到此时的时候，会在pPoint的指针地址上加z在类�?/span>Point3D中的偏移�?/span>8。由�?/span>pPoint的地址�?/span>0�Q�因此最�l?/span>offset的值是8�?/span>

&(pPoint->z)的语意是�?/span>pPoint中变�?/span>z的地址�Q?/span>pPoint的地址0�?/span>z的偏�U�量8�Q�，�q�不需要访�?/span>pPoint指向的内存。只要不讉K��非法的内存，�E�序��׃��会出错�?/span>

题目�Q�七�Q�：�q�行下列C++代码�Q�输��Z��么？

class A

{

public:

A()

{

Print();

}

virtual void Print()

{

printf("A is constructed.\n");

}

};

class B: public A

{

public:

B()

{

Print();

}

virtual void Print()

{

printf("B is constructed.\n");

}

};

int _tmain(int argc, _TCHAR* argv[])

{

A* pA = new B();

delete pA;

return 0;

}

�{�案�Q?/span>先后打印��Z��?/span>:A is constructed. B is constructed. 调用B的构造函数时�Q�先会调�?/span>B的基�c�d��A的构造函数。然后在A的构造函数里调用Print。由于此时实例的�c�d��B的部分还没有构造好�Q�本质上它只�?/span>A的一个实例，他的虚函数表指针指向的是�c�d��A的虚函数表。因此此时调用的Print�?/span>A::Print�Q�而不�?/span>B::Print。接着调用�c�d��B的构造函敎ͼ��q�调�?/span>Print。此时已�l�开始构�?/span>B�Q�因此此时调用的Print�?/span>B::Print�?/span>

同样是调用虚拟函�?/span>Print�Q�我们发现在�c�d��A的构造函��C��Q�调用的�?/span>A::Print�Q�在B的构造函��C��Q�调用的�?/span>B::Print。因此虚函数在构造函��C��Q�已�l�失��M��虚函数的动态绑定特性�?/span>

simplyzhao 2011-06-22 19:41 发表评论

simplyzhao — Fri, 17 Jun 2011 12:00:00 GMT

题目来源:
http://zhedahht.blog.163.com/blog/static/254111742010111112236313/

模拟�?

#include<stdio.h>
#define MAX_LEN 101

void
print_circle(int (*mtrx)[MAX_LEN], int leftup_x, int leftup_y, int rightdown_x, int rightdown_y)
{
    int i, j;
    if(leftup_x == rightdown_x) {
        for(j=leftup_y; j<=rightdown_y; j++)
            printf("%d\t", mtrx[leftup_x][j]);
        return;
    }
    if(leftup_y == rightdown_y) {
        for(i=leftup_x; i<=rightdown_x; i++)
            printf("%d\t", mtrx[i][leftup_y]);
        return;
    }

    for(i=leftup_y; i<rightdown_y; i++)
        printf("%d\t", mtrx[leftup_x][i]);
    for(j=leftup_x; j<rightdown_x; j++)
        printf("%d\t", mtrx[j][rightdown_y]);
    for(i=rightdown_y; i>leftup_y; i--)
        printf("%d\t", mtrx[rightdown_x][i]);
    for(j=rightdown_x; j>leftup_x; j--)
        printf("%d\t", mtrx[j][leftup_y]);
}

void
solve(int (*mtrx)[MAX_LEN], int width, int length)
{
    int lu_x, lu_y, rd_x, rd_y;
    lu_x = lu_y = 0;
    rd_x = width-1;
    rd_y = length-1;
    while(1) {
        if(lu_x>rd_x || lu_y>rd_y)
            break;
        print_circle(mtrx, lu_x, lu_y, rd_x, rd_y);
        ++lu_x;
        ++lu_y;
        --rd_x;
        --rd_y;
    }
}

int
main(int argc, char **argv)
{
    int i, j, length, width, matrix[MAX_LEN][MAX_LEN];
    scanf("%d %d", &width, &length);
    for(i=0; i<width; i++)
        for(j=0; j<length; j++)
            scanf("%d", matrix[i]+j);

    solve(matrix, width, length);
    return 0;
}

simplyzhao 2011-06-17 20:00 发表评论

simplyzhao — Fri, 17 Jun 2011 09:30:00 GMT

题目来源:
http://zhedahht.blog.163.com/blog/static/254111742011125100605/

题目�Q�写一个函敎ͼ�求两个整数的之和�Q�要求在函数体内不得使用�Q�、－�?#215;�?#247;�?/span>

分析�Q�这又是一道考察发散思维的很有意思的题目。当我们习以为常的东西被限制使用的时候，如何�H�破常规��L��考，��是解决�q�个问题的关键所在�?/span>

看到的这个题目，我的�W�一反应是傻��g��Q�四则运��都不能用，那还能用什么啊�Q�可是问题��L��要解决的�Q�只能打开思�\��L��考各�U�可能性。首先我们可以分析�h们是如何做十�q�制的加法的�Q�比如是如何得出5+17=22�q�个�l�果的。实际上�Q�我们可以分成三步的�Q�第一步只做各位相加不�q�位�Q�此时相加的�l�果�?/span>12�Q�个位数5�?/span>7相加不要�q�位�?/span>2�Q�十位数0�?/span>1相加�l�果�?/span>1�Q�；�W�二步做�q�位�Q?/span>5+7中有�q�位�Q�进位的值是10�Q�第三步把前面两个结果加��h��Q?/span>12+10的结果是22�Q�刚�?/span>5+17=22�?/span>

前面我们��在惻I��求两��C��和四则运��都不能用，那还能用什么啊�Q�对呀�Q�还能用什么呢�Q�对数字做运��，除了四则�q�算之外�Q�也��只剩下位运��了。位�q�算是针对二�q�制的，我们也就以二�q�制再来分析一下前面的三步走策略对二进制是不是也管用�?/span>

5的二�q�制�?/span>101�Q?/span>17的二�q�制10001。还是试着把计��分成三步：�W�一步各位相加但不计�q�位�Q�得到的�l�果�?/span>10100�Q�最后一位两个数都是1�Q�相加的�l�果是二�q�制�?/span>10。这一步不计进位，因此�l�果仍然�?/span>0�Q�；�W�二步记下进位。在�q�个例子中只在最后一位相加时产生一个进位，�l�果是二�q�制�?/span>10�Q�第三步把前两步的结果相加，得到的结果是10110�Q�正好是22。由此可见三步走的策略对二进制也是管用的�?/span>

接下来我们试着把二�q�制上的加法用位�q�算来替代。第一步不考虑�q�位�Q�对每一位相加�?/span>0�?/span>0�?/span> 1�?/span>1的结果都0�Q?/span>0�?/span>1�?/span>1�?/span>0的结果都�?/span>1。我们可以注意到�Q�这和异或的�l�果是一��L��。对异或而言�Q?/span>0�?/span>0�?/span>1�?/span>1异或的结果是0�Q��?/span>0�?/span>1�?/span>1�?/span>0的异或结果是1。接着考虑�W�二步进位，�?/span>0�?/span>0�?/span>0�?/span>1�?/span>1�?/span>0而言�Q�都不会产生�q�位�Q�只�?/span>1�?/span>1�Ӟ��会向前��生一个进位。此时我们可以想象成是两个数先做位与�q�算�Q�然后再向左�U�d��一位。只有两个数都是1的时候，位与得到的结果是1�Q�其余都�?/span>0。第三步把前两个步骤的结果相加。如果我们定义一个函�?/span>AddWithoutArithmetic�Q�第三步��q��当于输入前两步骤的结果来递归调用自己�?/p>

#include<stdio.h>

int
tricky_add(int arg1, int arg2)
{
    int a, b;
    a = arg1 ^ arg2; /* this is the result of arg1+arg2 without carry */
    b = arg1 & arg2;
    b <<= 1;

    if(b == 0)
        return a;
    else
        return tricky_add(a, b);
}

int
main(int argc, char **argv)
{
    int x, y;
    while(scanf("%d %d", &x, &y) != EOF) {
        printf("%d\n", tricky_add(x, y));
    }

    return 0;
}

simplyzhao 2011-06-17 17:30 发表评论

和�ؓn�q�箋正数序列

simplyzhao — Thu, 16 Jun 2011 11:33:00 GMT

题目来源:
http://zhedahht.blog.163.com/blog/static/25411174200732711051101/

题目�Q�输入一个正�?/span>n�Q�输出所有和�?/span>n�q�箋正数序列�?/span>

例如输入15�Q�由�?/span>1+2+3+4+5=4+5+6=7+8=15�Q�所以输�?/span>3个连�l�序�?/span>1-5�?/span>4-6�?/span>7-8�?/span>

分析�Q�这是网易的一道面试题�?/span>

�q�道题和本面试题�p�d��的第10�?/span>有些�c�M��。我们用两个�?/span>small�?/span>big分别表示序列的最��值和最大倹{��首先把small初始化�ؓ1�Q?/span>big初始化�ؓ2。如果从small�?/span>big的序列的和大�?/span>n的话�Q�我们向右移�?/span>small�Q�相当于从序列中��L��较小的数字。如果从small�?/span>big的序列的和小�?/span>n的话�Q�我们向右移�?/span>big�Q�相当于向序列中��d��big的下一个数字。一直到small�{�于(1+n)/2�Q�因为序列至��要有两个数字�?/span>

��Z��q�个思�\�Q�我们可以写出如下代码：

void PrintContinuousSequence(int small, int big);

/////////////////////////////////////////////////////////////////////////
// Find continuous sequence, whose sum is n
/////////////////////////////////////////////////////////////////////////
void FindContinuousSequence(int n)
{
      if(n < 3)
            return;

      int small = 1;
      int big = 2;
      int middle = (1 + n) / 2;
      int sum = small + big;

      while(small < middle)
      {
            // we are lucky and find the sequence
            if(sum == n)
                  PrintContinuousSequence(small, big);

            // if the current sum is greater than n,
            // move small forward
            while(sum > n)
            {
                  sum -= small;
                  small ++;

                  // we are lucky and find the sequence
                  if(sum == n)
                        PrintContinuousSequence(small, big);
            }

            // move big forward
            big ++;
            sum += big;
      }
}

/////////////////////////////////////////////////////////////////////////
// Print continuous sequence between small and big
/////////////////////////////////////////////////////////////////////////
void PrintContinuousSequence(int small, int big)
{
      for(int i = small; i <= big; ++ i)
            printf("%d ", i);

      printf("\n");
}

simplyzhao 2011-06-16 19:33 发表评论

simplyzhao — Sun, 12 Jun 2011 10:19:00 GMT

题目来源:
http://zhedahht.blog.163.com/blog/static/25411174200731844235261/

题目�Q�一个台阶��d��?/span>n�U�，如果一�ơ可以蟩1�U�，也可以蟩2�U�。求��d��有多��总蟩法，�q�分析算法的旉��复杂度�?/span>

分析�Q�这道题最�q�经常出玎ͼ�包括MicroStrategy�{�比较重视算法的公司都曾先后选用�q�个�q�道题作为面试题或者笔试题�?/span>

首先我们考虑最��单的情况。如果只�?/span>1�U�台�Ӟ��那显然只有一�U�蟩法。如果有2�U�台�Ӟ��那就有两�U�蟩的方法了�Q�一�U�是分两�ơ蟩�Q�每�ơ蟩1�U�；另外一�U�就是一�ơ蟩2�U��?/span>

现在我们再来讨论一般情��c��我们把n�U�台阶时的蟩法看成是n的函敎ͼ��Cؓf(n)。当n>2�Ӟ��W�一�ơ蟩的时候就有两�U�不同的选择�Q�一是第一�ơ只�?/span>1�U�，此时��x��数目�{�于后面剩下�?/span>n-1�U�台阶的��x��数目�Q�即�?/span>f(n-1)�Q�另外一�U�选择是第一�ơ蟩2�U�，此时��x��数目�{�于后面剩下�?/span>n-2�U�台阶的��x��数目�Q�即�?/span>f(n-2)。因�?/span>n�U�台阶时的不同蟩法的��L��f(n)=f(n-1)+(f-2)�?/span>

我们把上面的分析用一个公式�ȝ��如下�Q?/span>

        /  1                         n=1
f(n)=      2                          n=2
        \  f(n-1)+(f-2)               n>2

分析到这里，�怿�很多人都能看��是我们熟悉�?/span>Fibonacci序列�?/span>

simplyzhao 2011-06-12 18:19 发表评论

反�{链表循环与递归

simplyzhao — Sun, 12 Jun 2011 08:15:00 GMT

题目来源�Q?br />http://zhedahht.blog.163.com/blog/static/2541117420073471124487/

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

struct Node {
    char value;
    struct Node *next;
};

struct Node *
list_reverse(struct Node *head)
{
    struct Node *tmp, *cur, *pre = NULL;
    cur = head;
    while(cur) {
        tmp = cur->next;
        cur->next = pre;
        pre = cur;
        cur = tmp;
    }
    return pre;
}

struct Node *
list_reverse_recursive(struct Node *head)
{
    struct Node *rv;
    if(head && head->next) {
        rv = list_reverse_recursive(head->next);
        head->next->next = head;
        head->next = NULL;
        return rv;
    } else
        return head;
}

void
test_print(struct Node *head)
{
    while(head) {
        printf("%c\t", head->value);
        head = head->next;
    }
    printf("\n");
}

int
main(int argc, char **argv)
{
    struct Node d = {'d', NULL};
    struct Node c = {'c', &d};
    struct Node b = {'b', &c};
    struct Node a = {'a', &b};

    test_print(&a);

    struct Node *rev_first = list_reverse(&a);

    test_print(rev_first);

    struct Node *rev_second = list_reverse_recursive(rev_first);

    test_print(rev_second);

    return 0;
}

simplyzhao 2011-06-12 16:15 发表评论

simplyzhao — Fri, 10 Jun 2011 11:46:00 GMT

题目来源:
http://blog.163.com/prevBlogPerma.do?host=zhedahht&srl=25411174200731139971&mode=prev

#include<stdio.h>
#include<stdlib.h>

#include<string.h>
#include<limits.h>
/*
* #define INT_MAX 2147483647 　　
* #define INT_MIN (-INT_MAX-1)
*/

enum Status {
    Success,
    Fail
};
enum Status ret;
int negative;

int
Str2Int(const char *input)
{
    long long num = 0;
    negative = 0;
    ret = Fail;
    if(input == NULL)
        return num;

    const char *ptr = input;
    if(*ptr=='+' || *ptr=='-') {
        if(*ptr == '-')
            negative = 1;
        ++ptr;
    }
    while(*ptr) {
        if(!(*ptr>='0' && *ptr<='9'))
            return num;

        if((!negative && num>INT_MAX) || (negative && (-num)<INT_MIN))
            return num;

        num = num*10 + (*ptr-'0');
        ++ptr;
    }
    ret = Success;
    return num;
}

#define MAX_LEN 101

int
main(int argc, char **argv)
{
    int result;
    char value[MAX_LEN];
    while(scanf("%s", value) != EOF) {
        result = Str2Int(value);
        if(ret == Success)
            printf("%d\n", negative ? (-result) : result);
        else
            printf("Invalid\n");
    }

    return 0;
}

simplyzhao 2011-06-10 19:46 发表评论

带随机指针的链表复制

simplyzhao — Thu, 09 Jun 2011 03:35:00 GMT

题目来源�Q?br />http://zhedahht.blog.163.com/blog/static/254111742010819104710337/

题目�Q�有一个复杂链表，其结炚w��了有一�?font face="Calibri">m_pNext指针指向下一个结点外�Q�还有一�?/span>m_pSibling指向链表中的��M��l�点或�?/span>NULL。其�l�点�?/span>C++定义如下�Q?/span>

struct ComplexNode

{

int m_nValue;

ComplexNode* m_pNext;

ComplexNode* m_pSibling;

};

代码�Q?

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

struct Node {
    char value;

    struct Node *next;
    struct Node *random;
};

void test_print(struct Node *);

struct Node *
list_copy_with_random_pointer(struct Node *head)
{
    struct Node *tmp, *ptr, *ret;

    ptr = head;
    while(ptr != NULL) {
        tmp = (struct Node *)malloc(sizeof(struct Node));
        tmp->value = (ptr->value)-32; /* from lowercase to uppercase, just for testing */
        tmp->next = ptr->next;
        tmp->random = NULL;

        ptr->next = tmp;

        ptr = ptr->next->next;
    }

    ptr = head;
    while(ptr != NULL) {
        ptr->next->random = ptr->random==NULL ? NULL : ptr->random->next;

        ptr = ptr->next->next;
    }

    ptr = head;
    ret = (head==NULL ? NULL : (head->next));
    while(ptr != NULL) {
        tmp = ptr->next;
        ptr->next = ptr->next->next;
        tmp->next = ptr->next==NULL ? NULL : ptr->next->next;

        ptr = ptr->next;
    }

    return ret;
}

void
test_print(struct Node *head)
{
    while(head != NULL) {
        printf("%c: [%c, %c]\n", head->value, head->next==NULL?'-':head->next->value, head->random==NULL?'-':head->random->value);

        head = head->next;
    }
}

int
main(int argc, char **argv)
{
    struct Node d = {'d', NULL, NULL};
    struct Node c = {'c', &d, NULL};
    struct Node b = {'b', &c, NULL};
    struct Node a = {'a', &b, NULL};
    a.random = &c;
    d.random = &b;

    test_print(&a);

    struct Node *copy = list_copy_with_random_pointer(&a);

    printf("\n\n");
    test_print(&a);
    printf("\n\n");
    test_print(copy);

    return 0;
}

simplyzhao 2011-06-09 11:35 发表评论

simplyzhao — Wed, 01 Jun 2011 11:58:00 GMT

题目来源�Q?http://blog.163.com/prevBlogPerma.do?host=zhedahht&srl=2541117420072159363370&mode=prev

题目�Q�输入一颗二元查找树�Q�将该树转换为它的镜像，卛_��转换后的二元查找树中�Q�左子树的结炚w��大于叛_��树的�l�点。用递归和��@环两�U�方法完成树的镜像�{换�?/span>

例如输入�Q?/span>

     8
    / \
6      10
/\       /\
5 7    9   11

输出�Q?/span>

      8
    / \
10    6
/\      /\
11  9 7 5

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

struct Node {
    int value;
    struct Node *left;
    struct Node *right;
};

void
bst_preorder(struct Node *root)
{
    if(root == NULL)
        return;

    printf("%d\t", root->value);
    bst_preorder(root->left);
    bst_preorder(root->right);
}

void
bst_mirror_recursive(struct Node *root) /* easy */
{
    if(root == NULL)
        return;

    struct Node *ptr = root->left;
    root->left = root->right;
    root->right = ptr;

    bst_mirror_recursive(root->left);
    bst_mirror_recursive(root->right);
}

/* STACK : naive */
#define STACK_SIZE 101
struct Stack {
    void *data[STACK_SIZE];
    int top;
};

void
stack_pop(struct Stack *stack)
{
    if((stack->top) >= 0)
        --(stack->top);
}

void *
stack_top(struct Stack *stack)
{
    if((stack->top) >= 0)
        return stack->data[stack->top];

    return NULL;
}

void
stack_push(struct Stack *stack, void *entity)
{
    stack->data[++(stack->top)] = entity;
}

int
stack_isempty(struct Stack *stack)
{
    return (stack->top) < 0;
}

void
bst_mirror_nonrecursive(struct Node *root, struct Stack *aux_stack) /* stack used : good method */
{
    stack_push(aux_stack, root);
    while(!stack_isempty(aux_stack)) {
        struct Node *node = (struct Node *)stack_top(aux_stack);

        struct Node *ptr = node->left;
        node->left = node->right;
        node->right = ptr;

        stack_pop(aux_stack);
        if(node->left)
            stack_push(aux_stack, node->left);
        if(node->right)
            stack_push(aux_stack, node->right);
    }
}

int
main(int argc, char **argv)
{
    struct Node a = {5, NULL, NULL};
    struct Node b = {7, NULL, NULL};
    struct Node c = {9, NULL, NULL};
    struct Node d = {11, NULL, NULL};
    struct Node e = {6, &a, &b};
    struct Node f = {10, &c, &d};
    struct Node g = {8, &e, &f};

    bst_preorder(&g);
    printf("\n");
    bst_mirror_recursive(&g);
    bst_preorder(&g);
    printf("\n");

    bst_mirror_recursive(&g);
    bst_preorder(&g);
    printf("\n");
    struct Stack aux = {{0}, -1};
    bst_mirror_nonrecursive(&g, &aux);
    bst_preorder(&g);
    printf("\n");

    return 0;
}

simplyzhao 2011-06-01 19:58 发表评论

simplyzhao — Tue, 31 May 2011 08:58:00 GMT

题目来源: http://zhedahht.blog.163.com/blog/static/25411174201142733927831/

题目�Q?/span>输入一��二叉树的根�l�点�Q�判断该树是不是�q��二叉树。如果某二叉树中��L��l�点的左叛_��树的深度相差不超�q?/strong>1�Q�那么它��是一��^衡二叉树�?br />

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define MAX(a, b) ((a)>(b) ? (a) : (b))
#define ABS(x) ((x)>=0 ? (x) : ((x)*(-1)))

struct Node {
    void *data;
    struct Node *left, *right;
};

int
depth(struct Node *root)
{
    if(root == NULL)
        return 0;

    int ldepth = depth(root->left);
    int rdepth = depth(root->right);

    return MAX(ldepth, rdepth) + 1;
}

int
btree_isbalanced_naive(struct Node *root) /* return 1 if balanced, or return 0 */
{
    if(root == NULL)
        return 1;

    int ldepth = depth(root->left);
    int rdepth = depth(root->right);
    int diff = ldepth - rdepth;
    if(ABS(diff) > 1)
        return 0;

    return (btree_isbalanced_naive(root->left) && btree_isbalanced_naive(root->right));
}

int
btree_isbalanced(struct Node *root, int *depth)
{
    if(root == NULL) {
        *depth = 0;
        return 1;
    }

    int ldepth, rdepth, diff;
    if(!btree_isbalanced(root->left, &ldepth) || !btree_isbalanced(root->right, &rdepth))
        return 0;

    *depth = MAX(ldepth, rdepth) + 1;
    diff = ldepth - rdepth;
    return (ABS(diff)<=1 ? 1 : 0);
}

int
main(int argc, char **argv)
{
    struct Node a = {NULL, NULL, NULL};
    struct Node b = {NULL, NULL, NULL};
    struct Node c = {NULL, NULL ,NULL};
    struct Node d = {NULL, &b, NULL};
    struct Node e = {NULL, &a, &d};
    struct Node f = {NULL, NULL, NULL};
    struct Node g = {NULL, &e, &f};

    int ret1 = btree_isbalanced_naive(&g);
    printf("%s\n", ret1 ? "YES" : "NO");

    int dpth = 0;
    int ret2 = btree_isbalanced(&g, &dpth);
    printf("%s : %d\n", ret2 ? "YES" : "NO", dpth);
    return 0;
}

simplyzhao 2011-05-31 16:58 发表评论

设计包含min函数的栈

simplyzhao — Wed, 25 May 2011 07:49:00 GMT

题目出处: http://zhedahht.blog.163.com/blog/static/25411174200712895228171/

题目�Q�定义栈的数据结构，要求��d��一�?/span>min函数�Q�能够得到栈的最��元素。要求函�?/span>min�?/span>push以及pop的时间复杂度都是O(1)�?/span>
#include "StackWithMin.h"
#include<cstdio>
#include<cstdlib>
#include<cstring>

const int StackWithMin::MAX_SIZE ;

StackWithMin::StackWithMin() :
    top_index(-1), min_index(-1)
{
    printf("StackWithMin Constructor\n");
}

StackWithMin::~StackWithMin()
{
    printf("StackWithMin Destructor\n");
}

int
StackWithMin::top() const
{
    if(top_index == -1) {
        printf("top() failed: Stack Empty\n");
        return -1;
    }

    return stack[top_index];
}

int
StackWithMin::get_min() const
{
    if(min_index == -1) {
        printf("get_min() failed: Stack Empty\n");
        return -1;
    }

    return stack[min_index];
}

void
StackWithMin::push(int value)
{
    if(top_index == -1) { /* stack empty */
        stack[++top_index] = value;
        min_index = top_index;
        return;
    }
    stack[++top_index] = value;
    if(value < stack[min_index]) {
        index[top_index] = min_index;
        min_index = top_index;
    }
}

int
StackWithMin::pop()
{
    if(top_index == -1) {
        printf("pop() failed: Stack Empty\n");
        return -1;
    }
    int ret = stack[top_index];
    if(min_index == top_index)
        min_index = index[top_index];
    --top_index;
    return ret;
}

class StackWithMin
{
    public:
        StackWithMin();
        ~StackWithMin();
        int get_min() const;
        int top() const;
        void push(int value);
        int pop();
    private:
        static const int MAX_SIZE = 101;
        int top_index, min_index;
        int stack[MAX_SIZE];
        int index[MAX_SIZE];
};

simplyzhao 2011-05-25 15:49 发表评论

把二元查找树转变成排序的双向链表

simplyzhao — Mon, 23 May 2011 01:09:00 GMT
题目出处: http://zhedahht.blog.163.com/

题目�Q�输入一��二元查找树�Q�将该二元查找树转换成一个排序的双向链表。要求不能创��Z�Q何新的结点，只调整指针的指向�?
　　比如��二元查找树
                                            10
                                          /    \
                                        6       14
                                      / \     /　 \
                                   　4     8 12 　 16
转换成双向链�?/span>

4=6=8=10=12=14=16�?/span>

思�\�Q�递归�Q�在一时找不到递归的灵感的时候，多考虑考虑递归的参敎ͼ�有时更重要的是考虑递归的返回�?br />      每处理一个节点，首先获取左子树和叛_��树所�q�回的链表，然后拼接

代码�Q?
#include<stdio.h>
#include<string.h>
#include<stdlib.h>

/* Problem: Convert a binary search tree into a sorted linkedlist */
/* When it comes to Tree-Structure, recursion is always the most common solution.
   When designing recursion solution, should consider:
       1. the parameters
       2(important). the return object
*/

struct Node {
    int value;
    struct Node *left;
    struct Node *right;
};

struct Node *
BTree2List(struct Node *root)
{
    if(root == NULL)
        return NULL;
    struct Node *ret = NULL;

    /* Convert the left tree into a sorted linkedlist */
    struct Node *l_linkedlist = BTree2List(root->left);
    ret = l_linkedlist==NULL ? root : l_linkedlist;

    /* Convert the right tree into a sorted linkedlist */
    struct Node *r_linkedlist = BTree2List(root->right);
    while(l_linkedlist && l_linkedlist->right)
        l_linkedlist = l_linkedlist->right;

    /* Combine */
    if(l_linkedlist)
        l_linkedlist->right = root;
    root->left = l_linkedlist;
    root->right = r_linkedlist;
    if(r_linkedlist)
        r_linkedlist->left = root;

    return ret;
}

int main(int argc, char** argv)
{
    struct Node a = {4, NULL, NULL};
    struct Node b = {8, NULL, NULL};
    struct Node c = {12, NULL, NULL};
    struct Node d = {16, NULL, NULL};
    struct Node e = {6, NULL, &b};
    struct Node f = {14, &c, NULL};
    struct Node g = {10, &e, &f};

    struct Node *ret = BTree2List(&g);
    while(ret && ret->right) {
        ret = ret->right;
    }

    while(ret) {
        printf("%d\n", ret->value);
        ret = ret->left;
    }

    return 0;
}

simplyzhao 2011-05-23 09:09 发表评论

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从n个items中随机选择m个的问题

�z�牌���法的一�U�实�?/a>

扑克牌的��子

和�ؓn�q�箋正数序列

反�{链表 循环与递归

带随机指针的链表复制

设计包含min函数的栈

把二元查找树转变成排序的双向链表

�z�牌��法的一�U�实�?/a>

反�{链表循环与递归