123.买卖股票的最佳时机III
class Solution:
def maxProfit(self, prices: List[int]) -> int:
if len(prices) <= 1:
return 0
if len(prices) == 2:
return max(0, prices[1] - prices[0])
if len(prices) == 3:
return max(prices[2] - prices[1], prices[1]- prices[0], prices[2]- prices[0], 0)
# dp[i][0] 第 i 天不操作
# dp[i][1] 第 i 天第一次持有
# dp[i][2] 第 i 天第一次卖出
# dp[i][3] 第 i 天第二次持有
# dp[i][4] 第 i 天第二次卖出
dp = [[0] * 5 for _ in range(len(prices))]
dp[0][0] = 0
dp[0][1] = -prices[0]
dp[0][2] = 0
dp[0][3] = -prices[0]
dp[0][4] = 0
for i in range(1, len(prices)):
dp[i][0] = dp[i-1][0]
dp[i][1] = max(dp[i-1][1], dp[i-1][0] - prices[i])
dp[i][2] = max(dp[i-1][2], dp[i-1][1] + prices[i])
dp[i][3] = max(dp[i-1][3], dp[i-1][2] - prices[i])
dp[i][4] = max(dp[i-1][4], dp[i-1][3] + prices[i])
return dp[-1][4]
188.买卖股票的最佳时机IV
class Solution:
def maxProfit(self, k: int, prices: List[int]) -> int:
if len(prices) == 0:
return 0
dp = [[0] * (2*k + 1) for _ in range(len(prices))]
for i in range(1, 2*k, 2):
dp[0][i] = -prices[0]
for i in range(1, len(prices)):
for j in range(0, 2*k, 2):
dp[i][j+1] = max(dp[i-1][j+1], dp[i-1][j] - prices[i])
dp[i][j+2] = max(dp[i-1][j+2], dp[i-1][j+1] + prices[i])
return dp[-1][2*k]