題目：

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 as many transactions as you like (ie, buy one and sell one share of the stock multiple times) with the following restrictions:

• You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
• After you sell your stock, you cannot buy stock on next day. (ie, cooldown 1 day)

Example:

prices = [1, 2, 3, 0, 2] maxProfit = 3 transactions = [buy, sell, cooldown, buy, sell]

提示：

這道題可以用動(dòng)態(tài)規(guī)劃的思路解決。但是一開始想的時(shí)候總是抽象不出狀態(tài)轉(zhuǎn)移方程來，之后看到了一種用狀態(tài)機(jī)的思路，覺得很清晰，特此拿來分享，先看如下狀態(tài)轉(zhuǎn)移圖：

這里我們把狀態(tài)分成了三個(gè)，根據(jù)每個(gè)狀態(tài)的指向，我們可以得出下面的狀態(tài)轉(zhuǎn)移方程：

• s0[i] = max(s0[i-1], s2[i-1])
• s1[i] = max(s1[i-1], s0[i-1] - price[i])
• s2[i] = s1[i-1] + price[i]

這樣就清晰了很多。

代碼：

1 class Solution { 2 public: 3 int maxProfit(vector<int>& prices){ 4 if (prices.size() <= 1) return 0; 5 vector<int> s0(prices.size(), 0); 6 vector<int> s1(prices.size(), 0); 7 vector<int> s2(prices.size(), 0); 8 s1[0] = -prices[0]; 9 s0[0] = 0; 10 s2[0] = INT_MIN; 11 for (int i = 1; i < prices.size(); i++) { 12 s0[i] = max(s0[i - 1], s2[i - 1]); 13 s1[i] = max(s1[i - 1], s0[i - 1] - prices[i]); 14 s2[i] = s1[i - 1] + prices[i]; 15 } 16 return max(s0[prices.size() - 1], s2[prices.size() - 1]); 17 } 18 };

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轉(zhuǎn)載于:https://www.cnblogs.com/jdneo/p/5228004.html

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