Leetcode解题: Best time to buy and sell stock with cooldown (309)

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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]

解法1：DP O(N) Time + O(N) Space

这题主要要想到用两个数组来记录每一天的两种状态。一个是第i天持有股票，一个是第i天未持有股票。
用sell和buy代表两个数组。buy就是第i天持有股票的投资组合的最大市值。sell就是第i天未持有股票时的投资组合的最大市值。
那么初始状态sell[0] = 0, buy[0] = -prices[0]这是表示如果第一天持有股票的话一定是买入操作，那么需要花去prices[0]的钱。
那么每一天投资组合的演化可以得出下面的关系

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第i天未持股的最大市值要么是上一天1）持股2）未持股。
如果未持股则市值不变，如果持股那么因为第i天一定要卖掉，
所以如果上一天持股的话则今天的最大市值只能是buy[i-1]+Prices[i]
sell[i] = max(sell[i - 1], buy[i - 1] + prices[i])
同理，
buy[i] = max(buy[i - 1], sell[i - 2] - prices[i])

依此写出如下程序，如果要求的话可以对空间进一步优化成O(1)
C++

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class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if (prices.size() <= 1) {
            return 0;
        }
        int n = prices.size();
        vector<int> sell(n, 0);
        vector<int> buy(n, 0);
        buy[0] = -prices[0];
        for (int i = 1; i < n; ++i) {
            sell[i] = max(sell[i - 1], buy[i - 1] + prices[i]);
            buy[i] = max(buy[i - 1], (i > 1? sell[i - 2]:0) - prices[i]);
        }
        return sell[n - 1];
    }
};

解法1： 另外一种解释

以下的说明直接来自leetcode discussion

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1. Define States
To represent the decision at index i:
buy[i]: Max profit till index i. The series of transaction is ending with a buy.
sell[i]: Max profit till index i. The series of transaction is ending with a sell.
To clarify:
Till index i, the buy / sell action must happen and must be the last action. It may not happen at index i. It may happen at i - 1, i - 2, ... 0.
In the end n - 1, return sell[n - 1]. Apparently we cannot finally end up with a buy. In that case, we would rather take a rest at n - 1.
For special case no transaction at all, classify it as sell[i], so that in the end, we can still return sell[n - 1]. Thanks @alex153 @kennethliaoke @anshu2.
2. Define Recursion
buy[i]: To make a decision whether to buy at i, we either take a rest, by just using the old decision at i - 1, or sell at/before i - 2, then buy at i, We cannot sell at i - 1, then buy at i, because of cooldown.
sell[i]: To make a decision whether to sell at i, we either take a rest, by just using the old decision at i - 1, or buy at/before i - 1, then sell at i.
So we get the following formula:
buy[i] = Math.max(buy[i - 1], sell[i - 2] - prices[i]);   
sell[i] = Math.max(sell[i - 1], buy[i - 1] + prices[i]);
3. Optimize to O(1) Space
DP solution only depending on i - 1 and i - 2 can be optimized using O(1) space.
Let b2, b1, b0 represent buy[i - 2], buy[i - 1], buy[i]
Let s2, s1, s0 represent sell[i - 2], sell[i - 1], sell[i]
Then arrays turn into Fibonacci like recursion:
b0 = Math.max(b1, s2 - prices[i]);
s0 = Math.max(s1, b1 + prices[i]);
4. Write Code in 5 Minutes
First we define the initial states at i = 0:
We can buy. The max profit at i = 0 ending with a buy is -prices[0].
We cannot sell. The max profit at i = 0 ending with a sell is 0.

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public int maxProfit(int[] prices) {
    if(prices == null || prices.length <= 1) return 0;
    int b0 = -prices[0], b1 = b0;
    int s0 = 0, s1 = 0, s2 = 0;
    for(int i = 1; i < prices.length; i++) {
    	b0 = Math.max(b1, s2 - prices[i]);
    	s0 = Math.max(s1, b1 + prices[i]);
    	b1 = b0; s2 = s1; s1 = s0;
    }
    return s0;
}