每天的价格以Nrr数组arr []的形式给出N天，任务是找到在那些天之前必须出售股票的条件下，通过购买和出售股票所能获得的最大利润再次购买，第二天不能出售股票。 (即，至少休息一天)。
例子：
Input: arr[] = {2, 4, 5, 0, 2}
Output: 4
Explanation: Buy at cost 2 and sell at cost 4, profit = 2. Buy at cost 0 and sell at cost 2, profit = 2. Total profit = 4.
Input: arr[] = {2, 0, 5, 1, 8}
Output: 8
方法 ：
考虑三种状态： REST ， HOLD和SOLD 。
REST means not doing anything.
HOLD means holding stock (bought a stock and have not sold).
SOLD means state after selling the stock.
要达到REST状态，请不要执行任何操作。
要达到“持有”状态，请购买股票。
为了达到售出状态，销售为这首有需要购买的股票。
通过以上动作，制作了过渡图。
通过使用此过渡图，可以发现利润。
在第i天：
- rest [i]表示在第i天休息产生的最大利润。由于第i个一天休息日，股价可能已经卖了(I-1)个天或可能不会被出售。 rest [i] = max(rest [i-1]，sold [i-1])的值。
- 保持通过购买由[i]表示最大利润上的第i个天或通过在某一天的购买前的第i个和第一天休息我。因此， hold [i] = max(hold [i-1]，rest [i-1] + price [i])的值。
- sold [i]表示通过卖出第i天获得的最大利润。股票必须日日卖它在我之前已经买了一些日子。因此，sold [i] = hold [i-1] + price [i] 。
因此，最终答案将是
maximum of sold[n-1] and rest[n-1].
下面是上述方法的实现：
C++
// C++ program for the above problem
#include
using namespace std;
int maxProfit(int prices[], int n)
{
// If there is only one day
// for buying and selling
// no profit can be made
if (n <= 1)
return 0;
// Array to store Maxprofit by
// resting on given day
int rest[n] = { 0 };
// Array to store Maxprofit by
// buying or resting on the
// given day
int hold[n] = { 0 };
// Array to store Maxprofit by
// selling on given day
int sold[n] = { 0 };
// Initially there will 0 profit
rest[0] = 0;
// Buying on 1st day results
// in negative profit
hold[0] = -prices[0];
// zero profit since selling
// before buying isn't possible
sold[0] = 0;
for (int i = 1; i < n; i++) {
// max of profit on (i-1)th
// day by resting and profit
// on (i-1)th day by selling.
rest[i] = max(rest[i - 1],
sold[i - 1]);
// max of profit by resting
// on ith day and
// buying on ith day.
hold[i] = max(hold[i - 1],
rest[i - 1]
- prices[i]);
// max of profit by selling
// on ith day
sold[i] = hold[i - 1] + prices[i];
}
// maxprofit
return max(rest[n - 1],
sold[n - 1]);
}
// Driver Code
int main()
{
int price[] = { 2, 4,
5, 0, 2 };
int n = sizeof(price)
/ sizeof(price[0]);
cout << maxProfit(price, n)
<< endl;
return 0;
} Java
// Java program for the above problem
class GFG{
static int maxProfit(int prices[], int n)
{
// If there is only one day
// for buying and selling
// no profit can be made
if (n <= 1)
return 0;
// Array to store Maxprofit by
// resting on given day
int rest[] = new int[n];
// Array to store Maxprofit by
// buying or resting on the
// given day
int hold[] = new int[9];
// Array to store Maxprofit by
// selling on given day
int sold[] = new int[9];
// Initially there will 0 profit
rest[0] = 0;
// Buying on 1st day results
// in negative profit
hold[0] = -prices[0];
// Zero profit since selling
// before buying isn't possible
sold[0] = 0;
for(int i = 1; i < n; i++)
{
// max of profit on (i-1)th
// day by resting and profit
// on (i-1)th day by selling.
rest[i] = Math.max(rest[i - 1],
sold[i - 1]);
// max of profit by resting
// on ith day and
// buying on ith day.
hold[i] = Math.max(hold[i - 1],
rest[i - 1] -
prices[i]);
// max of profit by selling
// on ith day
sold[i] = hold[i - 1] + prices[i];
}
// maxprofit
return Math.max(rest[n - 1],
sold[n - 1]);
}
// Driver Code
public static void main(String[] args)
{
int price[] = { 2, 4, 5, 0, 2 };
int n = price.length;
System.out.print(maxProfit(price, n) + "\n");
}
}
// This code is contributed by amal kumar choubeyPython3
# Python3 program for the above problem
def maxProfit(prices, n):
# If there is only one day
# for buying and selling
# no profit can be made
if (n <= 1):
return 0
# Array to store Maxprofit by
# resting on given day
rest = [0] * n
# Array to store Maxprofit by
# buying or resting on the
# given day
hold = [0] * n
# Array to store Maxprofit by
# selling on given day
sold = [0] * n
# Initially there will 0 profit
rest[0] = 0
# Buying on 1st day results
# in negative profit
hold[0] = -prices[0]
# zero profit since selling
# before buying isn't possible
sold[0] = 0
for i in range(1, n):
# max of profit on (i-1)th
# day by resting and profit
# on (i-1)th day by selling.
rest[i] = max(rest[i - 1],
sold[i - 1])
# max of profit by resting
# on ith day and
# buying on ith day.
hold[i] = max(hold[i - 1],
rest[i - 1] -
prices[i])
# max of profit by selling
# on ith day
sold[i] = hold[i - 1] + prices[i]
# maxprofit
return max(rest[n - 1],
sold[n - 1])
# Driver Code
price = [ 2, 4, 5, 0, 2 ]
n = len(price)
print(maxProfit(price, n))
# This code is contributed by avanitrachhadiya2155C#
// C# program for the above problem
using System;
class GFG{
static int maxProfit(int[] prices, int n)
{
// If there is only one day
// for buying and selling
// no profit can be made
if (n <= 1)
return 0;
// Array to store Maxprofit by
// resting on given day
int[] rest = new int[n];
// Array to store Maxprofit by
// buying or resting on the
// given day
int[] hold = new int[9];
// Array to store Maxprofit by
// selling on given day
int[] sold = new int[9];
// Initially there will 0 profit
rest[0] = 0;
// Buying on 1st day results
// in negative profit
hold[0] = -prices[0];
// Zero profit since selling
// before buying isn't possible
sold[0] = 0;
for(int i = 1; i < n; i++)
{
// max of profit on (i-1)th
// day by resting and profit
// on (i-1)th day by selling.
rest[i] = Math.Max(rest[i - 1],
sold[i - 1]);
// max of profit by resting
// on ith day and
// buying on ith day.
hold[i] = Math.Max(hold[i - 1],
rest[i - 1] -
prices[i]);
// max of profit by selling
// on ith day
sold[i] = hold[i - 1] + prices[i];
}
// maxprofit
return Math.Max(rest[n - 1],
sold[n - 1]);
}
// Driver code
static void Main()
{
int[] price = { 2, 4, 5, 0, 2 };
int n = price.Length;
Console.WriteLine(maxProfit(price, n));
}
}
// This code is contributed by divyeshrabadiya07输出：
4
时间复杂度： O(N)
辅助空间： O(N)