数据结构与算法分析_数组的子序列问题

来源：互联网发布：homebrew安装mysql 编辑：程序博客网时间：2024/05/21 06:37
<span style="font-family:Arial;">/***2.17*求最小子序列和，最小正子序列和，最大子序列乘积**/#include <iostream>#include <algorithm>  using namespace std;#define INT_MIN -32768#define INT_MAX 32767//在里面有使用#define和const定义常量的比较，推荐使用const//#define LEN 10const int LEN = 10;/***最大子序列和**/int subMaxSum(int *a, int length){int thismax = 0, max=INT_MIN;for(int i=0; i<length; i++){thismax += a[i];if(thismax > max){max = thismax;}else if(thismax < 0){thismax = 0;}}return max;}/***最小子序列和*n*/int subMinSum(int a[], int length){int thismin = 0, min = INT_MAX;for(int i=0; i<length; i++){thismin += a[i];if(thismin < min){min = thismin;}else if(thismin > 0){thismin = 0;}}return min;}struct Node  {      int sum;      int xiabiao;        };  int cmp(const Node& t1,const Node& t2)  {      return t1.sum < t2.sum;} /***最小正子序列和*n*logn*/int positiveSubMinSum(int *data, int len){Node* temp = new Node[len];    temp[0].sum = data[0];    temp[0].xiabiao = 0;    for(int i=1;i<len;i++)    {      temp[i].sum = temp[i-1].sum+data[i];      temp[i].xiabiao = i;    }    //对temp.sum[]进行从小到大排序，sum[]中只有相邻的两个数才有可能 得到 最小正子序列和    sort(temp,temp+len,cmp);    int sum = INT_MAX;    for(int i=0;i<len-1;i++)    {      if(temp[i].xiabiao < temp[i+1].xiabiao)      {           if(temp[i+1].sum - temp[i].sum > 0 && temp[i+1].sum - temp[i].sum < sum)           sum = temp[i+1].sum - temp[i].sum;      }    }    delete temp;    temp=0;    return sum;  }void swap(int& a, int& b){ int temp = a; a = b; b = temp;}/***最大子序列乘积(同时也求出了最小的子序列乘积)*在找最大的值得时候，必须记录最小值，因为有负数存在，最小的数可能变成最大的数*/int mutiSubMax(int *a, int length){  int i;    int maxProduct = 1;    int minProduct = 1;    int maxCurrent = 1;    int minCurrent = 1;    for( i=0; i<length ;i++)    {        maxCurrent *= a[i];        minCurrent *= a[i];        if(maxCurrent > maxProduct)             maxProduct = maxCurrent;        if(minCurrent > maxProduct)            maxProduct = minCurrent;        if(maxCurrent < minProduct)            minProduct = maxCurrent;        if(minCurrent < minProduct)            minProduct = minCurrent;        //注意交换        if(minCurrent > maxCurrent)            swap(maxCurrent,minCurrent);        //这个必须在最后（防止为0的时候）        if(maxCurrent<1)            maxCurrent = 1;        //if(minCurrent>1)//这里不需要，因为通过交换即可，只需要一个        //  minCurrent =1;    }    return maxProduct;}int main(){int a[LEN] = {4,-1,5,-2,-1,2,6,-2,1,-3};cout<<"列表："<<endl;for(int i=0; i<10; i++){cout<<a[i]<<" ";}cout<<endl;cout<<"最大子序列和:"<<subMaxSum(a, LEN)<<endl;cout<<"最小子序列和:"<<subMinSum(a, LEN)<<endl;cout<<"最小正子序列和:"<<positiveSubMinSum(a, LEN)<<endl;cout<<"最大子序列乘积:"<<mutiSubMax(a, LEN)<<endl;return 0;}</span>
0 0