Say you have an array for which the ith 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]
买卖股票问题，这里多了一个冷却期，在卖出股票后的第二天不能交易，必须隔一天才能在买。
需要维护三个一维数组buy, sell，和rest。其中：
buy[i]表示在第i天之前最后一个操作是买，此时的最大收益。
sell[i]表示在第i天之前最后一个操作是卖，此时的最大收益。
rest[i]表示在第i天之前最后一个操作是冷冻期，此时的最大收益。
我们写出递推式为：
buy[i] = max(rest[i-1] - price, buy[i-1])
sell[i] = max(buy[i-1] + price, sell[i-1])
rest[i] = max(sell[i-1], buy[i-1], rest[i-1])
上述递推式很好的表示了在买之前有冷冻期，买之前要卖掉之前的股票。一个小技巧是如何保证[buy, rest, buy]的情况不会出现，这是由于buy[i] <= rest[i]， 即rest[i] = max(sell[i-1], rest[i-1])，这保证了[buy, rest, buy]不会出现。
另外，由于冷冻期的存在，可以得出rest[i] = sell[i-1]，这样，可以将上面三个递推式精简到两个：
buy[i] = max(sell[i-2] - price, buy[i-1])
sell[i] = max(buy[i-1] + price, sell[i-1])
做进一步优化，由于i只依赖于i-1和i-2，所以可以在O(1)的空间复杂度完成算法，
Java：
public int maxProfit(int[] prices) {
int sell = 0, prev_sell = 0, buy = Integer.MIN_VALUE, prev_buy;
for (int price : prices) {
prev_buy = buy;
buy = Math.max(prev_sell - price, prev_buy);
prev_sell = sell;
sell = Math.max(prev_buy + price, prev_sell);
}
return sell;
}
Python：
def maxProfit(self, prices):
if len(prices) < 2:
return 0
sell, buy, prev_sell, prev_buy = 0, -prices[0], 0, 0
for price in prices:
prev_buy = buy
buy = max(prev_sell - price, prev_buy)
prev_sell = sell
sell = max(prev_buy + price, prev_sell)
return sell
C++：
class Solution {
public:
int maxProfit(vector<int>& prices) {
int buy = INT_MIN, pre_buy = 0, sell = 0, pre_sell = 0;
for (int price : prices) {
pre_buy = buy;
buy = max(pre_sell - price, pre_buy);
pre_sell = sell;
sell = max(pre_buy + price, pre_sell);
}
return sell;
}
};
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