题目
You are given an integer array prices where prices[i] is the price of a given stock on the ith day.
Design an algorithm to find the maximum profit. You may complete at most k transactions.
Notice that you may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).
解题思路
如果k大于n/2，就变成第122题LeetCode122解答,同时也推荐先看第123题，只能进行两次交易的简化版LeetCode123解答
dp[i][0][1] 第一次买入
dp[i][1][0] 第一次卖出。
dp[i][j-1][1]：表示第i天，第j次买入后的累计最大利润
dp[i][j][0]：表示第i天，第j次卖出后的累计最大利润
dp[i][0][0] = 0
第j次买入： 第j次买入后保持不动 从第j-1次卖出后买入
dp[i][j-1][1] = max(dp[i-1][j-1][1], dp[i-1][j-1][0] - prices[i])
第j次卖出： 第j次卖出后保持不动 第j次买入后卖出
dp[i][j][0] = max(dp[i-1][j][0], dp[i-1][j-1][1] + prices[i])
因为第j次买入只和第j-1卖出相关，第j次卖出只和第j次买入相关，所以可以省去第一维
dp[j-1][1] = max(dp[j-1][1], dp[j-1][0] - prices[i])
dp[j][0] = max(dp[j][0], dp[j-1][1] + prices[i])
代码
class Solution:
def maxProfit(self, k: int, prices: List[int]) -> int:
if not prices:
return 0
n = len(prices)
if k > n // 2:
res = 0
for i in range(1, len(prices)):
if prices[i] > prices[i - 1]:
res += prices[i] - prices[i - 1]
return res
n = len(prices)
dp = [[0, 0] for _ in range(k + 1)]
for i in range(k + 1):
dp[i][1] = -prices[0]
for i in range(1, n):
for j in range(1, k + 1):
dp[j - 1][1] = max(dp[j - 1][1], dp[j - 1][0] - prices[i])
dp[j][0] = max(dp[j][0], dp[j - 1][1] + prices[i])
return dp[k][0]
