309. Best Time to Buy and Sell Stock with Cooldown

題目：

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]

連接： http://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-cooldown/

題解：

股票題又來啦，這應該是目前股票系列的最後一題。賣出以後有cooldown，而後求multi transaction的最大profit。第一印象就是dp，但每次dp的題目，轉移方程怎麼也寫很差，必定要好好增強。出這道題的dietpepsi在discuss裏也寫了他的思路和解法，你們都去看一看。不過我本身沒看懂....dp功力太差了， 反而是後面有一個哥們的state machine解法比較說得通。上面解法是based on一天只有一個操做，或買或賣或hold。也有一些理解爲能夠當天買賣的解法，列在了discuss裏。看來題目真的要指定得很是仔細，不然讀明白都很困難。

Time Complexity - O(n)， Space Complexity - O(n)

public class Solution {
    public int maxProfit(int[] prices) {
        if(prices == null || prices.length == 0) {
            return 0;
        }
        int len = prices.length;
        int[] buy = new int[len + 1];       // before i, for any sequence last action at i is going to be buy
        int[] sell = new int[len + 1];      // before i, for any sequence last action at i is going to be sell
        int[] cooldown = new int[len + 1];  // before i, for any sequence last action at i is going to be cooldown
        buy[0] = Integer.MIN_VALUE;
        
        for(int i = 1; i < len + 1; i++) {
            buy[i] = Math.max(buy[i - 1], cooldown[i - 1] - prices[i - 1]);     // must sell to get profit
            sell[i] = Math.max(buy[i - 1] + prices[i - 1], sell[i - 1]);
            cooldown[i] = Math.max(sell[i - 1], Math.max(buy[i - 1], cooldown[i - 1]));
        }
        
        return Math.max(buy[len], Math.max(sell[len], cooldown[len]));
    }
}

 

使用State machine的

public class Solution {
    public int maxProfit(int[] prices) {
        if(prices == null || prices.length < 2) {
            return 0;
        }
        int len = prices.length;
        int[] s0 = new int[len];                // to buy
        int[] s1 = new int[len];                // to sell
        int[] s2 = new int[len];                // to rest
        s0[0] = 0;
        s1[0] = -prices[0];
        s2[0] = 0;
        
        for(int i = 1; i < len; i++) {
            s0[i] = Math.max(s0[i - 1], s2[i - 1]);         
            s1[i] = Math.max(s1[i - 1], s0[i - 1] - prices[i]);
            s2[i] = s1[i - 1] + prices[i];
        }
        
        return Math.max(s0[len - 1], s2[len - 1]);      // hold and res
    }
}

 

有機會還要簡化space complexity， 要看一看yavinci的解析。

 

題外話：

今天剛發現leetcode提交解答的頁面加上了題目號，方便了很多，之前只是總目錄有題號。 這題跟個人狀況很像。如今公司的policy是買入股票必須hold 30天，並且不能夠買地產類股票...我以爲本身也沒接觸什麼數據，就作不了短線，真的很虧..

Reference:

https://leetcode.com/discuss/72030/share-my-dp-solution-by-state-machine-thinking

http://fujiaozhu.me/?p=725

http://bookshadow.com/weblog/2015/11/24/leetcode-best-time-to-buy-and-sell-stock-with-cooldown/

https://leetcode.com/discuss/71391/easiest-java-solution-with-explanations

http://www.cnblogs.com/grandyang/p/4997417.html

https://leetcode.com/discuss/71246/line-constant-space-complexity-solution-added-explanation

https://leetcode.com/discuss/73617/7-line-java-only-consider-sell-and-cooldown

https://leetcode.com/discuss/71354/share-my-thinking-process