【LeetCode】Best Time to Buy and Sell Stock IV

Best Time to Buy and Sell Stock IV

Say you have an array for which the i^th element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete at most k transactions.

Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

[Solution]

这题如果一开始不重视 k 的条件限制，很容易误以为是一道简单的贪心题。

大概会想，无非就是找出序列中 所有的上升区段，并用上升区段的最后一个数减第一个数（即此上升区段的跨度）作为其 value，最后再取 前 k 个最大的 values 求和得解。

典型的贪心思想，我一开始这样想写的代码如下：

 1 class Solution:
 2     # @return an integer as the maximum profit
 3     def maxProfit(self, k, prices):
 4         if prices == []:
 5             return 0
 6         down = prices[0]
 7         up = prices[0]
 8         ansList = []
 9         for i in range(1, len(prices)):
10             if prices[i-1] < prices[i]:
11                 up = prices[i]
12             elif prices[i] < prices[i-1]:
13                 ansList.append(up - down)
14                 down = prices[i]
15                 up = prices[i]
16         ansList.append(up - down)
17         ansList.sort(reverse=True)
18         if k < len(ansList):
19             ans = sum(ansList[:k])
20         else:
21             ans = sum(ansList)
22         return ans
23
24         

而且这贪心可以过：200 / 211 test cases passed.

但细想一下，正是因为有 k 的限制，其实 贪心 是不对的。

考虑这样一个样例：

【未完待续】