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 two transactions.
Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).
Input: [3,3,5,0,0,3,1,4]
Output: 6
Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.
Input: [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
engaging multiple transactions at the same time. You must sell before buying again.
Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.
给定一个股票每天的售价,如果你可以完成最多两次交易,求最大的收益。注意,你必须要卖出股票才能买入新的股票。
输入: [3, 3, 5, 0, 0, 3, 1, 4]
输出: 6
解释: 在第四天购入股票并在第六天卖出股票,收益为 3。然后在第七天购入股票,在第八天卖出股票,
收益为 3。总收益为 6,为最大收益。
这道题与Best Time to Buy and Sell Stock 及 Best Time to Buy and Sell Stock 2 不同,这两道题分别是完成最多一次交易和完成任意多次交易。而这道题,要求是最多完成两次交易。我们的难点即在于如何处理完成两次交易的情形。
我们利用与前两题类似的方法来做,但这道题由于可以完成最多两次交易,故我们需要四个数组 dp1
、dp2
、dp3
、dp4
:
dp1[i]
表示第 i+1
天未完成任何交易并持有股票的最大收益,dp1[0]
初始化为 -prices[0]
dp2[i]
表示第 i+1
天完成一次任何交易并不持有股票的最大收益,dp2[0]
初始化为 0
dp3[i]
表示第 i+1
天完成一次交易并持有股票的最大收益,dp3[0]
初始化为 -INF