算法: 买卖股票的最佳时机 IV 188. Best Time to Buy and Sell Stock IV

Posted 2022-03-10 AI架构师易筋

188. Best Time to Buy and Sell Stock IV

You are given an integer array prices where prices[i] is the price of a given stock on the ith day, and an integer k.

Find the maximum profit you can achieve. You may complete at most k transactions.

Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).

Example 1:

Input: k = 2, prices = [2,4,1]
Output: 2
Explanation: Buy on day 1 (price = 2) and sell on day 2 (price = 4), profit = 4-2 = 2.

Example 2:

Input: k = 2, prices = [3,2,6,5,0,3]
Output: 7
Explanation: Buy on day 2 (price = 2) and sell on day 3 (price = 6), profit = 6-2 = 4. Then buy on day 5 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.

Constraints:

• 0 <= k <= 100
• 0 <= prices.length <= 1000
• 0 <= prices[i] <= 1000

动态规划解法

这是 O(kn)，显然不是最优的。我将关键变量命名为局部利润和全局利润，以使事情更容易理解. 处理分了3部分

1. 如果n少于2次交易，return 0
2. 如果交易次数k >= n /2, 所以只要现在大于上一次，就做交易，累计和即可。
3. 当前的k层的交易max(profit, 0)，加上k-1层交易的最大收益，则为当前的要是做交易最大收益；跟当前层放弃这次交易的最大收益做对比。

https://youtu.be/oDhu5uGq_ic

class Solution:
    def maxProfit(self, k: int, prices: List[int]) -> int:
        n = len(prices)
        if n < 2:
            return 0
        # k is big enougth to cover all ramps.
        if k >= n / 2:
            return sum(i - j for i, j in zip(prices[1:], prices[:-1]) if i - j > 0)
        globalMax = [[0] * n for _ in range(k+1)]
        for r in range(1, k+1):
            localMax = [0] * n
            for c in range(1, n):
                profit = prices[c] - prices[c - 1]
                localMax[c] = max(
                	# We have made max profit with (r - 1) transations in (c - 1) days.
	                # 1. For the last transation, we buy stock on day (c - 1) and sell it on day j.
	                # 2. For the last transation, we buy stock on day c and
	                # sell it on the same day, so we have 0 profit, apparently we do not have to add it.
                    globalMax[r-1][c-1] + max(profit, 0),
                    # We have made profit in (c - 1) days.
	                # We want to cancel the day (c - 1) sale and sell it on day c.
                    localMax[c-1] + profit
                )
                globalMax[r][c] = max(globalMax[r][c-1], localMax[c])
        return globalMax[k][-1
]