编程珠玑（第二版）系列算法实现

最近开始攻读编程珠玑，这里将实现的部分程序汇集起来，使用的IDE是VS 2010.

第一章：

（1）内存足够情况下，使用C/C++标准库函数实现数组排序

C库函数实现：

#include "stdafx.h"
#include <iostream>
#include <cstdlib>
using namespace std;

int intcomp(const void* x,const void* y)
{
    return *(int*)x-*(int*)y;
}
int _tmain(int argc, _TCHAR* argv[])
{
    const int n = 5;
    int a[n];
    int i = 0;
    while(scanf("%d", &a[i]) != EOF)
        ++i;
    qsort(a, n, sizeof(int), intcomp);
    for(i = 0; i < n; i++)
    {
        printf("%d
",a[i]);
    }
    printf("
");

    system("pause");
    return 0;
}

C++库函数实现：

#include "stdafx.h"
#include <iostream>
#include <set>
using namespace std;

int _tmain(int argc, _TCHAR* argv[])
{
    set<int> s;
    int i;
    while(cin>>i)
        s.insert(i);

    for(set<int>::iterator iter = s.begin(); iter != s.end(); iter++)
        cout<<*iter<<endl;

    system("pause");
    return 0;
}

（2）使用位逻辑运算（如与、或、移位）来实现位向量并用此实现位排序 

#include "stdafx.h"
#include <iostream>
using namespace std;

const int bitsPerWord = 32;
const int shift = 5;
const int mask = 0x1F;
const int N = 10000000;
int a[1+N/bitsPerWord];

void set(int i)
{
    a[i>>shift] |= (1<<(i & mask));
}
void clr(int i)
{
    a[i>>shift] &= ~(1<<(i & mask));
}
int test(int i)
{
    return a[i>>shift] & (1<<(i & mask));
}

int _tmain(int argc, _TCHAR* argv[])
{
    int i;
    for(i = 0; i < N; i++)
        clr(i);

    while(cin>>i)
        set(i);
    for(i = 0; i < N; i++)
        if(test(i))
            cout<<i<<endl;

    system("pause");
    return 0;
}

 第二章：

（3）实现2.3节讲述的类似杂技的向量旋转代码

#include "stdafx.h"
#include <iostream>
using namespace std;

int _tmain(int argc, _TCHAR* argv[])
{
    const int n = 9;
    char x[n+1] = "cbfdgehai";
    cout<<x<<endl;
    
    int rotdist = 3;
    for(int i = 0; i < rotdist; i++)
    {
        int t = x[i];
        int j = i;
        while(1)
        {                    
            int k = j + rotdist;
            if(k >= n)
                k -= n;
            if(k == i)
                break;
            x[j] = x[k];
            j = k;
            x[j] = t;
        }
    }

    cout<<x<<endl;
    system("pause");
    return 0;
}

 第八章：

四种算法实现求一个数组中的最大连续子向量和，分别为简单算法(O(n^3))，平方算法(O(n^2))，分治算法(O（nlogn))，扫描算法(O(n))。C++实现如下：

#include "stdafx.h"
#include <iostream>
using namespace std;
#include <time.h>

void maxsum1(int a[], int n)
{
    int start = clock();
    if(n < 1)
        return;
    int maxsofar = 0;
    for(int i = 0;i < n; i++)
    {
        for(int j = i; j < n; j++)
        {
            int sum = 0;
            for(int k = i; k <= j;k++)
            {
                sum += a[k];
                maxsofar = max(sum,maxsofar);
            }
        }
    }
    int time = clock()-start;
    cout<<"maxsum:"<<maxsofar<<"
Time:"<<time<<endl;
}

void maxsum2a(int a[], int n)
{
    int start = clock();
    if(n < 1)
        return;
    int maxsofar = 0;
    for(int i = 0; i< n; i++)
    {
        int sum = 0;
        for(int j = i; j < n; j++)
        {
            sum += a[j];
            maxsofar = max(maxsofar, sum);
        }
    }
    int time = clock()-start;
    cout<<"maxsum2a:"<<maxsofar<<"
Time:"<<time<<endl;
}

void maxsum2b(int a[], const int n)
{
    int start = clock();
    if(n < 1)
        return;
    const int LEN = 100;
    int cumarr[LEN] = { 0 };
    for(int i = 0; i < n; i++)
    {
        cumarr[i+1] = cumarr[i]+a[i];
    }
    int maxsofar = 0;
    for(int i = 0; i < n; i++)
    {
        for(int j = i; j < n; j++)
        {
            int sum = cumarr[j+1]-cumarr[i];
            maxsofar = max(maxsofar, sum);
        }
    }
    int time = clock()-start;
    cout<<"maxsum2b:"<<maxsofar<<"
Time:"<<time<<endl;
}

int maxsum3Rev(int a[], int l, int r)
{
    if(l > r)
        return 0;
    if(l == r)
        return a[l];
    int m = (l+r)/2;
    int lmax = 0;
    int rmax = 0;
    int sum = 0;
    for(int i = m; i >= l; i--)
    {
        sum += a[i];
        lmax = max(lmax, sum);
    }
    sum = 0;
    for(int j = m+1; j <= r; j++)
    {
        sum += a[j];
        rmax = max(rmax, sum);
    }
    int maxsum = max(maxsum3Rev(a,l,m), maxsum3Rev(a,m+1,r));
    maxsum = max(maxsum,lmax+rmax);

    return maxsum;
}

void maxsum3(int a[], int n)
{
    int start = clock();
    if(n < 1)
        return;
    int maxsum = maxsum3Rev(a,0,n-1);
    int time = clock()-start;
    cout<<"maxsum3:"<<maxsum<<"
Time:"<<time<<endl;
}

void maxsum4(int a[],int n)
{
    int start = clock();
    int maxsofar = 0;
    int maxendinghere = 0;
    for(int i = 0; i < n; i++)
    {
        maxendinghere = max(maxendinghere + a[i], 0);
        maxsofar = max(maxsofar, maxendinghere);
    }
    int time = clock()-start;
    cout<<"maxsum4:"<<maxsofar<<"
Time:"<<time<<endl;
}

int _tmain(int argc, _TCHAR* argv[])
{
    int a[] = {3, 2, 1, -13, 10, -2, -1,6};
    maxsum1(a,8);
    maxsum2a(a,8);
    maxsum2b(a,8);
    maxsum3(a,8);
    maxsum4(a,8);

    system("pause");
    return 0;
}

延伸：如果要找的是和最接近0的连续子向量，该如何做呢？此时应当在算法2b的基础上得到数组cumarr(cumarr[i] = x[0]+x[1]+...+x[i])，然后对数组cumarr排序，相邻元素做差，找出cumarr中最接近的两个元素。基本原理：如果cumarr[i-1] = cumarr[j]，那么x[i..j]的和为0。