通过最多买卖两次股份获得最大利润

在日常股票交易中，买主在早上购买股票并在同一天出售。如果允许交易者一天最多进行两笔交易，而第二笔交易只能在第一笔交易完成后才能开始(卖出->买入->卖出->买入)。给定全天的股价，找出股票交易者可以赚到的最大利润。

例子：

Input:   price[] = {10, 22, 5, 75, 65, 80}
Output:  87
Trader earns 87 as sum of 12, 75 
Buy at 10, sell at 22, 
Buy at 5 and sell at 80

Input:   price[] = {2, 30, 15, 10, 8, 25, 80}
Output:  100
Trader earns 100 as sum of 28 and 72
Buy at price 2, sell at 30, buy at 8 and sell at 80

Input:   price[] = {100, 30, 15, 10, 8, 25, 80};
Output:  72
Buy at price 8 and sell at 80.

Input:   price[] = {90, 80, 70, 60, 50}
Output:  0
Not possible to earn.

一个简单的解决方案是考虑每个索引“ i”并执行以下操作

Max profit with at most two transactions =
       MAX {max profit with one transaction and subarray price[0..i] +
            max profit with one transaction and subarray price[i+1..n-1]  }
i varies from 0 to n-1.

可以使用以下O(n)算法计算一次交易的最大可能金额
两个元素之间的最大差异，使得较大的元素出现在较小的数字之后
上述简单解决方案的时间复杂度为O(n ² )。

我们可以使用以下有效解决方案来实现O(n)。想法是存储每个子阵列的最大可能利润，并在接下来的两个阶段中解决问题。

1)创建一个表profit [0..n-1]并将表中的所有值初始化为0。
2)从右到左遍历price []并更新利润[i]，以使利润[i]将一次交易可获得的最大利润存储在子数组价格[i..n-1]中
3)从左到右遍历价格[]并更新利润[i]，以使利润[i]存储最大利润，从而使利润[i]包含子数组价格[0..i]中两次交易的最大可实现利润。
4)回报利润[n-1]

要执行步骤2，我们需要从右到左跟踪最大价格，而要执行步骤3，我们需要从左到右跟踪最小价格。为什么我们要反向走?这个想法是为了节省空间，在第三步中，我们为两个目的使用相同的数组，最多1个事务，最大2个事务。在迭代i之后，数组Profit [0..i]包含2个事务的最大利润，而Profit [i + 1..n-1]包含2个事务的利润。

以下是上述想法的实现。

C++

// C++ program to find maximum
// possible profit with at most
// two transactions
#include 
using namespace std;
 
// Returns maximum profit with
// two transactions on a given
// list of stock prices, price[0..n-1]
int maxProfit(int price[], int n)
{
    // Create profit array and
    // initialize it as 0
    int* profit = new int[n];
    for (int i = 0; i < n; i++)
        profit[i] = 0;
 
    /* Get the maximum profit with
       only one transaction
       allowed. After this loop,
       profit[i] contains maximum
       profit from price[i..n-1]
       using at most one trans. */
    int max_price = price[n - 1];
    for (int i = n - 2; i >= 0; i--) {
        // max_price has maximum
        // of price[i..n-1]
        if (price[i] > max_price)
            max_price = price[i];
 
        // we can get profit[i] by taking maximum of:
        // a) previous maximum, i.e., profit[i+1]
        // b) profit by buying at price[i] and selling at
        //    max_price
        profit[i]
            = max(profit[i + 1], max_price - price[i]);
    }
 
    /* Get the maximum profit with two transactions allowed
       After this loop, profit[n-1] contains the result */
    int min_price = price[0];
    for (int i = 1; i < n; i++) {
        // min_price is minimum price in price[0..i]
        if (price[i] < min_price)
            min_price = price[i];
 
        // Maximum profit is maximum of:
        // a) previous maximum, i.e., profit[i-1]
        // b) (Buy, Sell) at (min_price, price[i]) and add
        //    profit of other trans. stored in profit[i]
        profit[i] = max(profit[i - 1],
                        profit[i] + (price[i] - min_price));
    }
    int result = profit[n - 1];
 
    delete[] profit; // To avoid memory leak
 
    return result;
}
 
// Driver code
int main()
{
    int price[] = { 2, 30, 15, 10, 8, 25, 80 };
    int n = sizeof(price) / sizeof(price[0]);
    cout << "Maximum Profit = " << maxProfit(price, n);
    return 0;
}

Java

class Profit {
    // Returns maximum profit
    // with two transactions on a
    // given list of stock prices,
    // price[0..n-1]
    static int maxProfit(int price[], int n)
    {
        // Create profit array
        // and initialize it as 0
        int profit[] = new int[n];
        for (int i = 0; i < n; i++)
            profit[i] = 0;
 
        /* Get the maximum profit
           with only one transaction
           allowed. After this loop,
           profit[i] contains
           maximum profit from
           price[i..n-1] using at most
           one trans. */
        int max_price = price[n - 1];
        for (int i = n - 2; i >= 0; i--) {
            // max_price has maximum
            // of price[i..n-1]
            if (price[i] > max_price)
                max_price = price[i];
 
            // we can get profit[i]
            // by taking maximum of:
            // a) previous maximum,
            // i.e., profit[i+1]
            // b) profit by buying
            // at price[i] and selling
            // at
            //    max_price
            profit[i] = Math.max(profit[i + 1],
                                 max_price - price[i]);
        }
 
        /* Get the maximum profit
           with two transactions allowed
           After this loop, profit[n-1]
           contains the result
         */
        int min_price = price[0];
        for (int i = 1; i < n; i++) {
            // min_price is minimum
            // price in price[0..i]
            if (price[i] < min_price)
                min_price = price[i];
 
            // Maximum profit is maximum of:
            // a) previous maximum, i.e., profit[i-1]
            // b) (Buy, Sell) at (min_price, price[i]) and
            // add
            // profit of other trans.
            // stored in profit[i]
            profit[i] = Math.max(
                profit[i - 1],
                profit[i] + (price[i] - min_price));
        }
        int result = profit[n - 1];
        return result;
    }
 
    // Driver Code
    public static void main(String args[])
    {
        int price[] = { 2, 30, 15, 10, 8, 25, 80 };
        int n = price.length;
        System.out.println("Maximum Profit = "
                           + maxProfit(price, n));
    }
 
} /* This code is contributed by Rajat Mishra */

Python

# Returns maximum profit with
# two transactions on a given
# list of stock prices price[0..n-1]
 
 
def maxProfit(price, n):
 
    # Create profit array and initialize it as 0
    profit = [0]*n
 
    # Get the maximum profit
    # with only one transaction
    # allowed. After this loop,
    # profit[i] contains maximum
    # profit from price[i..n-1]
    # using at most one trans.
    max_price = price[n-1]
 
    for i in range(n-2, 0, -1):
 
        if price[i] > max_price:
            max_price = price[i]
 
        # we can get profit[i] by
        # taking maximum of:
        # a) previous maximum,
        # i.e., profit[i+1]
        # b) profit by buying at
        # price[i] and selling at
        #    max_price
        profit[i] = max(profit[i+1], max_price - price[i])
 
    # Get the maximum profit
    # with two transactions allowed
    # After this loop, profit[n-1]
    # contains the result
    min_price = price[0]
 
    for i in range(1, n):
 
        if price[i] < min_price:
            min_price = price[i]
 
        # Maximum profit is maximum of:
        # a) previous maximum,
        # i.e., profit[i-1]
        # b) (Buy, Sell) at
        # (min_price, A[i]) and add
        #  profit of other trans.
        # stored in profit[i]
        profit[i] = max(profit[i-1], profit[i]+(price[i]-min_price))
 
    result = profit[n-1]
 
    return result
 
 
# Driver function
price = [2, 30, 15, 10, 8, 25, 80]
print "Maximum profit is", maxProfit(price, len(price))
 
# This code is contributed by __Devesh Agrawal__

C#

// C# program to find maximum possible profit
// with at most two transactions
using System;
 
class GFG {
 
    // Returns maximum profit with two
    // transactions on a given list of
    // stock prices, price[0..n-1]
    static int maxProfit(int[] price, int n)
    {
 
        // Create profit array and initialize
        // it as 0
        int[] profit = new int[n];
        for (int i = 0; i < n; i++)
            profit[i] = 0;
 
        /* Get the maximum profit with only
        one transaction allowed. After this
        loop, profit[i] contains maximum
        profit from price[i..n-1] using at
        most one trans. */
        int max_price = price[n - 1];
 
        for (int i = n - 2; i >= 0; i--) {
 
            // max_price has maximum of
            // price[i..n-1]
            if (price[i] > max_price)
                max_price = price[i];
 
            // we can get profit[i] by taking
            // maximum of:
            // a) previous maximum, i.e.,
            // profit[i+1]
            // b) profit by buying at price[i]
            // and selling at max_price
            profit[i] = Math.Max(profit[i + 1],
                                 max_price - price[i]);
        }
 
        /* Get the maximum profit with two
        transactions allowed After this loop,
        profit[n-1] contains the result */
        int min_price = price[0];
 
        for (int i = 1; i < n; i++) {
 
            // min_price is minimum price in
            // price[0..i]
            if (price[i] < min_price)
                min_price = price[i];
 
            // Maximum profit is maximum of:
            // a) previous maximum, i.e.,
            // profit[i-1]
            // b) (Buy, Sell) at (min_price,
            // price[i]) and add profit of
            // other trans. stored in
            // profit[i]
            profit[i] = Math.Max(
                profit[i - 1],
                profit[i] + (price[i] - min_price));
        }
        int result = profit[n - 1];
 
        return result;
    }
 
    // Driver code
    public static void Main()
    {
        int[] price = { 2, 30, 15, 10, 8, 25, 80 };
        int n = price.Length;
 
        Console.Write("Maximum Profit = "
                      + maxProfit(price, n));
    }
}
 
// This code is contributed by nitin mittal.

PHP

= 0; $i--)
    {
        // max_price has maximum
        // of price[i..n-1]
        if ($price[$i] > $max_price)
            $max_price = $price[$i];
 
        // we can get profit[i] by
        // taking maximum of:
        // a) previous maximum,
        //    i.e., profit[i+1]
        // b) profit by buying at
        // price[i] and selling at
        // max_price
        if($profit[$i + 1] >
           $max_price-$price[$i])
        $profit[$i] = $profit[$i + 1];
        else
        $profit[$i] = $max_price -
                      $price[$i];
    }
 
    // Get the maximum profit with
    // two transactions allowed.
    // After this loop, profit[n-1]
    // contains the result
    $min_price = $price[0];
    for ($i = 1; $i < $n; $i++)
    {
        // min_price is minimum
        // price in price[0..i]
        if ($price[$i] < $min_price)
            $min_price = $price[$i];
 
        // Maximum profit is maximum of:
        // a) previous maximum,
        //    i.e., profit[i-1]
        // b) (Buy, Sell) at (min_price,
        //     price[i]) and add
        // profit of other trans.
        // stored in profit[i]
        $profit[$i] = max($profit[$i - 1],
                          $profit[$i] +
                         ($price[$i] - $min_price));
    }
    $result = $profit[$n - 1];
    return $result;
}
 
// Driver Code
$price = array(2, 30, 15, 10,
               8, 25, 80);
$n = sizeof($price);
echo "Maximum Profit = ".
      maxProfit($price, $n);
     
// This code is contributed
// by Arnab Kundu
?>

输出

Maximum Profit = 100

上述解决方案的时间复杂度为O(n)。
算法范例：动态编程

还有一种使用Valley-Peak方法计算此问题的方法，即获取可变利润并将其初始化为零，然后在初始位置值较大时从第(i + 1)个位置遍历price []数组比之前的值将其添加到可变利润中。

C++

#include 
using namespace std;
 
int main()
{
    int price[] = { 2, 30, 15, 10, 8, 25, 80 };
    int n = 7;
   
    // adding array
    int profit = 0;
   
    // Initializing variable
    // valley-peak approach
    /*
                       80
                       /
        30            /
       /  \          25
      /    15       /
     /      \      /
    2        10   /
               \ /
                8
     */
    for (int i = 1; i < n; i++)
    {
       
        // traversing through array from (i+1)th
        // position
        int sub = price[i] - price[i - 1];
        if (sub > 0)
            profit += sub;
    }
 
    cout << "Maximum Profit=" << profit;
    return 0;
}
 
// This code is contributed by RohitOberoi.

Java

import java.io.*;
import java.util.*;
 
class GFG {
    public static void main(String[] args)
    {
        Scanner sc = new Scanner(System.in);
 
        int price[] = { 2, 30, 15, 10, 8, 25, 80 };
        // adding array
        int profit = 0;
        // Initializing variable
        // valley-peak approach
        /*
                           80
                           /
            30            /
           /  \          25
          /    15       /
         /      \      /
        2        10   /
                   \ /
                    8
         */
        for (int i = 1; i < price.length; i++) {
            // traversing through array from (i+1)th
            // position
            int sub = price[i] - price[i - 1];
            if (sub > 0)
                profit += sub;
        }
        System.out.print("Maximum Profit=" + profit);
    }
}

C＃

using System;
using System.Collections.Generic;
class GFG {
 
    static void Main()
    {
 
        int[] price = { 2, 30, 15, 10, 8, 25, 80 };
 
        // adding array
        int profit = 0;
 
        // Initializing variable
        // valley-peak approach
        /*
                           80
                           /
            30            /
           /  \          25
          /    15       /
         /      \      /
        2        10   /
                   \ /
                    8
         */
        for (int i = 1; i < price.Length; i++) {
 
            // traversing through array from (i+1)th
            // position
            int sub = price[i] - price[i - 1];
            if (sub > 0)
                profit += sub;
        }
        Console.WriteLine("Maximum Profit=" + profit);
    }
}
 
// This code is contributed by divyeshrabadiya07

输出

Maximum Profit=100

时间和空间复杂度分别为O(n)和O(1)。