股票买入卖出以最大化利润的 C 程序

有效的方法：如果我们只允许买卖一次，那么我们可以使用以下算法。两个元素之间的最大差异。在这里，我们可以多次买卖。
下面是这个问题的算法。

1. 找到局部最小值并将其存储为起始索引。如果不存在，则返回。
2. 找到局部最大值。并将其存储为结束索引。如果我们到达结尾，则将结尾设置为结束索引。
3. 更新解决方案（增加买卖对的计数）
4. 如果没有到达终点，重复上述步骤。

C

// Program to find best buying and 
// selling days
#include 
  
// Solution structure
struct Interval 
{
    int buy;
    int sell;
};
  
// This function finds the buy and sell 
// schedule for maximum profit
void stockBuySell(int price[], int n)
{
    // Prices must be given for at 
    // least two days
    if (n == 1)
        return;
  
    // Count of solution pairs
    int count = 0; 
  
    // Solution vector
    Interval sol[n / 2 + 1];
  
    // Traverse through given price array
    int i = 0;
    while (i < n - 1) 
    {
        // Find Local Minima. Note that the 
        // limit is (n-2) as we are comparing 
        // present element to the next element.
        while ((i < n - 1) && 
               (price[i + 1] <= price[i]))
            i++;
  
        // If we reached the end, break as no 
        // further solution possible
        if (i == n - 1)
            break;
  
        // Store the index of minima
        sol[count].buy = i++;
  
        // Find Local Maxima.  Note that the 
        // limit is (n-1) as we are comparing 
        // to previous element
        while ((i < n) && 
               (price[i] >= price[i - 1]))
            i++;
  
        // Store the index of maxima
        sol[count].sell = i - 1;
  
        // Increment count of buy/sell pairs
        count++;
    }
  
    // Print solution
    if (count == 0)
        printf(
        "There is no day when buying the stock will make profitn");
    else 
    {
        for (int i = 0; i < count; i++)
            printf(
            "Buy on day: %dt Sell on day: %dn", sol[i].buy, sol[i].sell);
    }
  
    return;
}
  
// Driver code
int main()
{
    // Stock prices on consecutive days
    int price[] = {100, 180, 260, 
                   310, 40, 535, 695};
    int n = sizeof(price) / sizeof(price[0]);
  
    // Function call
    stockBuySell(price, n);
  
    return 0;
}

输出：

Buy on day: 0     Sell on day: 3
Buy on day: 4     Sell on day: 6

时间复杂度：外循环一直运行到我变成 n-1。内部两个循环在每次迭代中增加 I 的值。所以整体时间复杂度是 O(n)

请参阅完整的股票买卖以最大化利润的文章了解更多详情！