本文概述在股票交易中, 买方在未来的某个日期购买股票并出售。给定股票价格为n天, 交易者最多可以进行k笔交易, 而新交易只能在上一次交易完成后才能开始, 请找出股票交易者可以赚取的最大利润。
例子：
Input:
Price = [10, 22, 5, 75, 65, 80]
K = 2
Output:87
Trader earns 87 as sum of 12 and 75
Buy at price 10, sell at 22, buy at
5 and sell at 80Input:
Price = [12, 14, 17, 10, 14, 13, 12, 15]
K = 3
Output:12
Trader earns 12 as the sum of 5, 4 and 3
Buy at price 12, sell at 17, buy at 10
and sell at 14 and buy at 12 and sell
at 15
Input:
Price = [100, 30, 15, 10, 8, 25, 80]
K = 3
Output:72
Only one transaction. Buy at price 8
and sell at 80.Input:
Price = [90, 80, 70, 60, 50]
K = 1
Output:0
Not possible to earn.有多种版本的问题。如果只允许我们买卖一次, 则可以使用两个元素之间的最大差额算法。如果我们最多只能进行2笔交易, 则可以按照讨论的方法进行这里。如果允许我们进行任意次数的买卖, 则可以遵循讨论的方法这里.
推荐：请在"实践首先, 在继续解决方案之前。在这篇文章中, 我们只允许进行最多k笔交易。该问题可以通过使用动态编程来解决。
让利润[t] [i]表示直到第i天(包括第i天)最多使用t笔交易获得的最大利润。那么关系是：
利润[t] [i] =最大值(利润[t] [i-1], 最大值(价格[i] –价格[j] +利润[t-1] [j]))
对于范围[0, i-1]中的所有j
利润[t] [i]最高为–
- Profit [t] [i-1], 表示在第i天没有进行任何交易。
- 在第i天卖出的最大利润。为了在第i天卖出股票, 我们需要在[0, i – 1]天中的任何一天购买股票。如果我们在第j天买入股票, 然后在第i天卖出, 则最大获利将为价格[i] –价格[j] +利润[t-1] [j], 其中j在0到i-1之间变化。在这里, 利润[t-1] [j]最好, 直到第j天, 我们只需减少一笔交易即可完成。
C ++
// C++ program to find out maximum profit by
// buying and selling a share atmost k times
// given stock price of n days
#include <
climits>
#include <
iostream>
using namespace std;
// Function to find out maximum profit by buying
// &
selling a share atmost k times given stock
// price of n days
int maxProfit( int price[], int n, int k)
{
// table to store results of subproblems
// profit[t][i] stores maximum profit using
// atmost t transactions up to day i (including
// day i)
int profit[k + 1][n + 1];
// For day 0, you can't earn money
// irrespective of how many times you trade
for ( int i = 0;
i <
= k;
i++)
profit[i][0] = 0;
// profit is 0 if we don't do any transation
// (i.e. k =0)
for ( int j = 0;
j <
= n;
j++)
profit[0][j] = 0;
// fill the table in bottom-up fashion
for ( int i = 1;
i <
= k;
i++) {
for ( int j = 1;
j <
n;
j++) {
int max_so_far = INT_MIN;
for ( int m = 0;
m <
j;
m++)
max_so_far = max(max_so_far, price[j] - price[m] + profit[i - 1][m]);
profit[i][j] = max(profit[i][j - 1], max_so_far);
}
}return profit[k][n - 1];
}// Driver code
int main()
{
int k = 2;
int price[] = { 10, 22, 5, 75, 65, 80 };
int n = sizeof (price) / sizeof (price[0]);
cout <
<
"Maximum profit is: "
<
<
maxProfit(price, n, k);
return 0;
}Java
// Java program to find out maximum
// profit by buying and selling a
// share atmost k times given
// stock price of n daysclass GFG {// Function to find out
// maximum profit by
// buying &
selling a
// share atmost k times
// given stock price of n days
static int maxProfit( int [] price, int n, int k)
{// table to store results
// of subproblems
// profit[t][i] stores
// maximum profit using
// atmost t transactions up
// to day i (including day i)
int [][] profit = new int [k + 1 ][n + 1 ];
// For day 0, you can't
// earn money irrespective
// of how many times you trade
for ( int i = 0 ;
i <
= k;
i++)
profit[i][ 0 ] = 0 ;
// profit is 0 if we don't
// do any transation
// (i.e. k =0)
for ( int j = 0 ;
j <
= n;
j++)
profit[ 0 ][j] = 0 ;
// fill the table in
// bottom-up fashion
for ( int i = 1 ;
i <
= k;
i++)
{
for ( int j = 1 ;
j <
n;
j++)
{
int max_so_far = 0 ;
for ( int m = 0 ;
m <
j;
m++)
max_so_far = Math.max(max_so_far, price[j] -
price[m] + profit[i - 1 ][m]);
profit[i][j] = Math.max(profit[i] [j - 1 ], max_so_far);
}
}return profit[k][n - 1 ];
}// Driver code
public static void main(String []args)
{
int k = 2 ;
int [] price = { 10 , 22 , 5 , 75 , 65 , 80 };
int n = price.length;
System.out.println( "Maximum profit is: " +
maxProfit(price, n, k));
}
}// This code is contributed by Anshul Aggarwal.Python3
# Python program to maximize the profit
# by doing at most k transactions
# given stock prices for n days# Function to find out maximum profit by
# buying &
selling a share atmost k times
# given stock price of n days
def maxProfit(prices, n, k):# Bottom-up DP approach
profit = [[ 0 for i in range (k + 1 )]
for j in range (n)]# Profit is zero for the first
# day and for zero transactions
for i in range ( 1 , n):for j in range ( 1 , k + 1 ):
max_so_far = 0for l in range (i):
max_so_far = max (max_so_far, prices[i] -
prices[l] + profit[l][j - 1 ])profit[i][j] = max (profit[i - 1 ][j], max_so_far)return profit[n - 1 ][k]# Driver code
k = 2
prices = [ 10 , 22 , 5 , 75 , 65 , 80 ]
n = len (prices)print ( "Maximum profit is:" , maxProfit(prices, n, k))# This code is contributed by vaibhav29498C#
// C# program to find out maximum profit by
// buying and selling a share atmost k times
// given stock price of n days
using System;
class GFG {// Function to find out maximum profit by
// buying &
selling/ a share atmost k times
// given stock price of n days
static int maxProfit( int [] price, int n, int k)
{
// table to store results of subproblems
// profit[t][i] stores maximum profit using atmost
// t transactions up to day i (including day i)
int [, ] profit = new int [k + 1, n + 1];
// For day 0, you can't earn money
// irrespective of how many times you trade
for ( int i = 0;
i <
= k;
i++)
profit[i, 0] = 0;
// profit is 0 if we don't do any transation
// (i.e. k =0)
for ( int j = 0;
j <
= n;
j++)
profit[0, j] = 0;
// fill the table in bottom-up fashion
for ( int i = 1;
i <
= k;
i++) {
for ( int j = 1;
j <
n;
j++) {
int max_so_far = 0;
for ( int m = 0;
m <
j;
m++)
max_so_far = Math.Max(max_so_far, price[j] -
price[m] + profit[i - 1, m]);
profit[i, j] = Math.Max(profit[i, j - 1], max_so_far);
}
}return profit[k, n - 1];
}// Driver code to test above
public static void Main()
{
int k = 2;
int [] price = { 10, 22, 5, 75, 65, 80 };
int n = price.Length;
Console.Write( "Maximum profit is: " +
maxProfit(price, n, k));
}
}// This code is contributed by Sam007的PHP
<
?php
// PHP program to find out maximum profit by
// buying and selling a share atmost k times
// given stock price of n days// Function to find out maximum profit by buying
// &
selling a share atmost k times given stock
// price of n days
function maxProfit( $price , $n , $k )
{// table to store results of subproblems
// profit[t][i] stores maximum profit using
// atmost t transactions up to day i (including
// day i)
$profit [ $k + 1][ $n + 1] = 0;
// For day 0, you can't earn money
// irrespective of how many times you trade
for ( $i = 0;
$i <
= $k ;
$i ++)
$profit [ $i ][0] = 0;
// profit is 0 if we don't
// do any transation
// (i.e. k =0)
for ( $j = 0;
$j <
= $n ;
$j ++)
$profit [0][ $j ] = 0;
// fill the table in
// bottom-up fashion
for ( $i = 1;
$i <
= $k ;
$i ++) {
for ( $j = 1;
$j <
$n ;
$j ++) {
$max_so_far = PHP_INT_MIN;
for ( $m = 0;
$m <
$j ;
$m ++)
$max_so_far = max( $max_so_far , $price [ $j ] - $price [ $m ] +
$profit [ $i - 1][ $m ]);
$profit [ $i ][ $j ] = max( $profit [ $i ][ $j - 1], $max_so_far );
}
}return $profit [ $k ][ $n - 1];
}// Driver code
$k = 2;
$price = array (10, 22, 5, 75, 65, 80 );
$n = sizeof( $price );
echo "Maximum profit is: " . maxProfit( $price , $n , $k );
// This is contributed by mits
?>
输出：
Maximum profit is: 87优化的解决方案：
上述解决方案的时间复杂度为O(k.n
2
)。如果我们能够计算出在固定时间的第i天卖出股票所获得的最大利润, 则可以减少该收益。
利润[t] [i] =最大值(利润[t] [i-1], 最大值(价格[i] –价格[j] +利润[t-1] [j])))
对于范围[0, i-1]中的所有j
如果我们仔细注意,
max(价格[i] –价格[j] +利润[t-1] [j])
对于范围[0, i-1]中的所有j
可以改写成
=价格[i] +最大值(利润[t-1] [j] –价格[j])
对于范围[0, i-1]中的所有j
=价格[i] +最大值(prevDiff, 利润[t-1] [i-1] –价格[i-1])
其中prevDiff为max(利润[t-1] [j] –价格[j])
对于范围[0, i-2]中的所有j
因此, 如果我们已经为范围[0, i-2]中的所有j计算了max(profit [t-1] [j] – price [j]), 则可以在恒定时间内为j = i – 1进行计算。换句话说, 我们不必再往回看[0, i-1]范围就可以找到购买的最佳日期。我们可以使用下面的修正关系确定恒定时间。
利润[t] [i] =最大值(利润[t] [i-1], 价格[i] +最大值(prevDiff, 利润[t-1] [i-1] –价格[i-1])
其中prevDiff是范围[0, i-2]中所有j的max(profit [t-1] [j] – price [j])
以下是其优化的实现–
C ++
// C++ program to find out maximum profit by buying
// and/ selling a share atmost k times given stock
// price of n days
#include <
climits>
#include <
iostream>
using namespace std;
// Function to find out maximum profit by buying &
// selling/ a share atmost k times given stock price
// of n days
int maxProfit( int price[], int n, int k)
{
// table to store results of subproblems
// profit[t][i] stores maximum profit using atmost
// t transactions up to day i (including day i)
int profit[k + 1][n + 1];
// For day 0, you can't earn money
// irrespective of how many times you trade
for ( int i = 0;
i <
= k;
i++)
profit[i][0] = 0;
// profit is 0 if we don't do any transation
// (i.e. k =0)
for ( int j = 0;
j <
= n;
j++)
profit[0][j] = 0;
// fill the table in bottom-up fashion
for ( int i = 1;
i <
= k;
i++) {
int prevDiff = INT_MIN;
for ( int j = 1;
j <
n;
j++) {
prevDiff = max(prevDiff, profit[i - 1][j - 1] - price[j - 1]);
profit[i][j] = max(profit[i][j - 1], price[j] + prevDiff);
}
}return profit[k][n - 1];
}// Driver Code
int main()
{
int k = 3;
int price[] = { 12, 14, 17, 10, 14, 13, 12, 15 };
int n = sizeof (price) / sizeof (price[0]);
cout <
<
"Maximum profit is: "
<
<
maxProfit(price, n, k);
return 0;
}Java
// Java program to find out maximum
// profit by buying and selling a
// share atmost k times given stock
// price of n days
import java.io.*;
class GFG {// Function to find out maximum profit by
// buying &
selling/ a share atmost k times
// given stock price of n days
static int maxProfit( int price[], int n, int k)
{// table to store results of subproblems
// profit[t][i] stores maximum profit
// using atmost t transactions up to day
// i (including day i)
int profit[][] = new int [k + 1 ][ n + 1 ];
// For day 0, you can't earn money
// irrespective of how many times you trade
for ( int i = 0 ;
i <
= k;
i++)
profit[i][ 0 ] = 0 ;
// profit is 0 if we don't do any
// transation (i.e. k =0)
for ( int j = 0 ;
j <
= n;
j++)
profit[ 0 ][j] = 0 ;
// fill the table in bottom-up fashion
for ( int i = 1 ;
i <
= k;
i++)
{
int prevDiff = Integer.MIN_VALUE;
for ( int j = 1 ;
j <
n;
j++)
{
prevDiff = Math.max(prevDiff, profit[i - 1 ][j - 1 ] -
price[j - 1 ]);
profit[i][j] = Math.max(profit[i][j - 1 ], price[j] + prevDiff);
}
}return profit[k][n - 1 ];
}// Driver code
public static void main (String[] args)
{
int k = 3 ;
int price[] = { 12 , 14 , 17 , 10 , 14 , 13 , 12 , 15 };
int n = price.length;
System.out.println( "Maximum profit is: " +
maxProfit(price, n, k));
}
} // This code is contributed by Sam007Python3
# Python3 program to find out maximum
# profit by buying and/ selling a share
# at most k times given the stock price
# of n days # Function to find out maximum profit
# by buying &
selling/ a share atmost
# k times given stock price of n days
def maxProfit(price, n, k): # Table to store results of subproblems
# profit[t][i] stores maximum profit
# using atmost t transactions up to
# day i (including day i)
profit = [[ 0 for i in range (n + 1 )]
for j in range (k + 1 )] # Fill the table in bottom-up fashion
for i in range ( 1 , k + 1 ):
prevDiff = float ( '-inf' )for j in range ( 1 , n):
prevDiff = max (prevDiff, profit[i - 1 ][j - 1 ] -
price[j - 1 ])
profit[i][j] = max (profit[i][j - 1 ], price[j] + prevDiff) return profit[k][n - 1 ] # Driver Code
if __name__ = = "__main__" :k = 3
price = [ 12 , 14 , 17 , 10 , 14 , 13 , 12 , 15 ]
n = len (price) print ( "Maximum profit is:" , maxProfit(price, n, k)) # This code is contributed
# by Rituraj JainC#
// C# program to find out maximum profit by buying
// and selling a share atmost k times given stock
// price of n days
using System;
class GFG {// Function to find out maximum profit by
// buying &
selling/ a share atmost k times
// given stock price of n days
static int maxProfit( int [] price, int n, int k)
{
// table to store results of subproblems
// profit[t][i] stores maximum profit using atmost
// t transactions up to day i (including day i)
int [, ] profit = new int [k + 1, n + 1];
// For day 0, you can't earn money
// irrespective of how many times you trade
for ( int i = 0;
i <
= k;
i++)
profit[i, 0] = 0;
// profit is 0 if we don't do any transation
// (i.e. k =0)
for ( int j = 0;
j <
= n;
j++)
profit[0, j] = 0;
// fill the table in bottom-up fashion
for ( int i = 1;
i <
= k;
i++) {
int prevDiff = int .MinValue;
for ( int j = 1;
j <
n;
j++) {
prevDiff = Math.Max(prevDiff, profit[i - 1, j - 1] - price[j - 1]);
profit[i, j] = Math.Max(profit[i, j - 1], price[j] + prevDiff);
}
}return profit[k, n - 1];
}// Driver code to test above
public static void Main()
{
int k = 3;
int [] price = {12, 14, 17, 10, 14, 13, 12, 15};
int n = price.Length;
Console.Write( "Maximum profit is: " +
maxProfit(price, n, k));
}
}// This code is contributed by Sam007的PHP
<
?php
// PHP program to find out maximum
// profit by buying and selling a
// share atmost k times given stock
// price of n days// Function to find out maximum
// profit by buying &
selling a
// share atmost k times given
// stock price of n days
function maxProfit( $price , $n , $k )
{// table to store results
// of subproblems profit[t][i]
// stores maximum profit using
// atmost t transactions up to
// day i (including day i)
$profit [ $k + 1][ $n + 1]=0;
// For day 0, you can't
// earn money irrespective
// of how many times you trade
for ( $i = 0;
$i <
= $k ;
$i ++)
$profit [ $i ][0] = 0;
// profit is 0 if we don't
// do any transation
// (i.e. k =0)
for ( $j = 0;
$j <
= $n ;
$j ++)
$profit [0][ $j ] = 0;
// fill the table in
// bottom-up fashion
$prevDiff = NULL;
for ( $i = 1;
$i <
= $k ;
$i ++) {
for ( $j = 1;
$j <
$n ;
$j ++) {
$prevDiff = max( $prevDiff , $profit [ $i - 1][ $j - 1] -
$price [ $j - 1]);
$profit [ $i ][ $j ] = max( $profit [ $i ][ $j - 1], $price [ $j ] + $prevDiff );
}
}
return $profit [ $k ][ $n - 1];
}// Driver Code
$k = 3;
$price = array (12, 14, 17, 10, 14, 13, 12, 15);
$n = sizeof( $price );
echo "Maximum profit is: "
, maxProfit( $price , $n , $k );
// This code is contributed by nitin mittal
?>
输出：
Maximum profit is: 12上述解决方案的时间复杂度为O(kn), 空间复杂度为O(nk)。当我们使用上一个事务的结果时, 空间复杂度可以进一步降低为O(n)。但是为了使文章易于阅读, 我们使用了O(kn)空间。
【算法设计（通过最多买卖k次股票获得最大利润）】本文作者：阿迪亚·戈尔(Aditya Goel)。如果发现任何不正确的地方, 或者想分享有关上述主题的更多信息, 请发表评论。
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