[编程之美]子数组的最大乘积

方法一

使用动态规划逐步求解

#include

using namespace std;

int max_product(int a[], int n) {

  int *s = new int[n+1];

  int *t = new int[n+1];

  int *p = new int[n+1];

  s[0] = 1;  // s[i]表示前i个元素的乘积

  t[n] = 1;  // t[i]表示后n-i个元素的乘积, p[i]表示除第i个元素之外其他N-1个元素的乘积

  for (int i = 1; i <= n; i++) {

    s[i] = s[i-1]*a[i-1];

  }

  for (int i = n; i >= 1; i--) {

    t[i-1] = t[i]*a[i-1];

  }

  for (int i = 1; i < n; i++) {

    p[i] = s[i-1]*t[i];

  }

  int max_index = 1;

  for (int i = 1; i < n; i++) {

    if (p[max_index] < p[i])

      max_index = i;

  }

  cout << "最大值为除去a[" << max_index << "]外的元素" << endl;

  return p[max_index];

}

int main() {

int a[10] = {1, 2, 3, -4, -6, 5, -8, 4, -2, -1};

  cout << max_product(a, 10) << endl;

  return 0;

}

方法二

根据正数负数和0的个数

#include

#include

using namespace std;

void max_product(int a[], int n) {

  int positive_num = 0;

  int min_positive = INT_MAX;

  int negative_num = 0;

  int max_negative = INT_MIN;

  int zero_num = 0;

  for (int i = 0; i < n; i++) {

    if (a[i] > 0) {

      ++positive_num;

      if (a[i] < min_positive)

        min_positive = a[i];

    } else if (a[i] < 0) {

      ++negative_num;

      if (a[i] > max_negative)

        max_negative = a[i];

    } else {

      zero_num++;

    }

      ++positive_num;

      if (a[i] < min_positive)

        min_positive = a[i];

    } else if (a[i] < 0) {

      ++negative_num;

      if (a[i] > max_negative)

        max_negative = a[i];

    } else {

      zero_num++;

    }

  int negative_num = 0;

  int max_negative = INT_MIN;

  int zero_num = 0;

  for (int i = 0; i < n; i++) {

    if (a[i] > 0) {

      ++positive_num;

      if (a[i] < min_positive)

        min_positive = a[i];

    } else if (a[i] < 0) {

      ++negative_num;

      if (a[i] > max_negative)

        max_negative = a[i];

    } else {

      ++zero_num;

    }

  } 

  

  if (zero_num > 1) {

    cout << "不管去掉哪个数结果都是0" << endl;

  } else if (zero_num == 1) {

    if (negative_num%2 == 0) {

      // 有偶数个负数

      cout << "去掉0" << endl;

    } else {

      //有奇数个负数，去掉绝对值最小的负数

      cout << "去掉" << max_negative << endl;

    }

  } else {

    if (negative_num%2 == 0) {

      // 有偶数个负数,去掉绝对值最小的正数

      cout << "去掉" << min_positive << endl;

    } else {

      //有奇数个负数，去掉绝对值最小的负数

      cout << "去掉" << max_negative << endl;

    }

  }

}

int main() {

  int a[5] = {1, 2, 3, 4, -6};

  max_product(a, 5);

  return 0;

}