编程珠玑 Pearls(8)

问题

输入具有n个浮点数的向量x,输出是输入向量的任意连续子向量的最大和。

8.2 .两个平方算法

int maxSum1(int arr[],int sz){    int i,j,maxsofar,sum;    maxsofar = 0;    sum = 0;    for(i = 0;i < sz;i++)    {        sum = 0;        for(j = i ;j < sz;j++)        {            sum = sum + arr[j];            maxsofar = max(sum,maxsofar);        }    }    return maxsofar;}//n*nint maxSum2(int arr[],int sz){    int i,j,maxsofar,sum;    int *arraux = (int*)malloc (sizeof(int) * (sz + 1));    arraux[0] = 0;    for(i = 1;i <= sz;i++)    {        arraux[i] = arraux[i-1] + arr[i - 1];    }    maxsofar = 0;    for(i = 1;i <= sz;i++)    {        for(j = i ;j <= sz;j++)        {            sum = arraux[j] - arraux[i - 1];            maxsofar = max(sum,maxsofar);        }    }    return maxsofar;}

8.3 .分治算法

//n*log nint maxSum3(int arr[],int sz,int lo,int up){    if(lo > up) return 0;    if(lo == up) return max(0,arr[lo]);    int m = lo + (up - lo) / 2;    int lmax,sum,rmax,i;    lmax = sum = 0;    for(i = m;i >= lo ; i--)    {        sum += arr[i];        lmax = max(lmax,sum);    }    rmax = sum = 0;    for(i = m + 1;i <= up; i++)    {        sum += arr[i];        rmax = max(rmax,sum);    }    return max(lmax+rmax,max(maxSum3(arr,sz,lo,m),maxSum3(arr,sz,m+1,up)));    }

8.4 .扫描算法

int maxSum4(int arr[],int sz){    int maxsofar = 0,maxendinghere = 0 ,i;    for(i = 0;i < sz;i++)    {        maxendinghere = max(maxendinghere+arr[i],0);        maxsofar = max(maxsofar,maxendinghere);    }    return maxsofar;}

测试程序

//#define LOCAL#include<iostream>#include<stdio.h>#include<stdlib.h>using namespace std;const int maxn=50001;int num[maxn],N;int cnt=0;void Init(){    cin >> N;     for(int i = 0;i < N;i++)    {        cin >> num[i];    }}int main(void){#ifndef LOCAL    freopen("1049.in","r",stdin);    freopen("1049.out","w",stdout);#endif    Init();    cout << maxSum1(num,N) << endl;    cout << maxSum2(num,N) << endl;    cout << maxSum3(num,N,0,N) << endl;    cout << maxSum4(num,N) << endl;    return 0;}