LeetCode -Best Time to Buy and Sell Stock with Cooldown

LeetCode -Best Time to Buy and Sell Stock with Cooldown

题目描述

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]

用一个数组表示股票每天的价格，数组的第i个数表示股票在第i天的价格。 如果某天卖了股票，那么第二天不能买股票，有一天的冷冻期。

算法解释

状态机想法（discuss 大神）

其中

s0[i] = max(s0[i - 1], s2[i - 1]); // Stay at s0, or rest from s2
s1[i] = max(s1[i - 1], s0[i - 1] - prices[i]); // Stay at s1, or buy from s0
s2[i] = s1[i - 1] + prices[i]; // Only one way from s1

---

s0[0] = 0; // At the start, you don’t have any stock if you just rest
s1[0] = -prices[0]; // After buy, you should have -prices[0] profit. Be positive!
s2[0] = 0;

---

S1：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。

S2：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。

S0：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。

---

代码

class Solution {
public:
    int maxProfit(vector<int>& prices){
        if (prices.size() <= 1) return 0;
        vector<int> s0(prices.size(), 0);
        vector<int> s1(prices.size(), 0);
        vector<int> s2(prices.size(), 0);
        s1[0] = -prices[0];
        s0[0] = 0;
        s2[0] = INT_MIN;
        for (int i = 1; i < prices.size(); i++) {
            s0[i] = max(s0[i - 1], s2[i - 1]);
            s1[i] = max(s1[i - 1], s0[i - 1] - prices[i]);
            s2[i] = s1[i - 1] + prices[i];
        }
        return max(s0[prices.size() - 1], s2[prices.size() - 1]);
    }
};

对意思进行理解优化后

int maxProfit(vector<int>& prices) {
        if (prices.size() < 2) return 0;
        int rest = 0, buy = -prices[0], sell = 0;
        for (int i = 1; i < prices.size(); ++i) {
            int last_sell = sell;
            sell = buy + prices[i];
            buy = max(rest - prices[i], buy);
            rest = max(rest, last_sell);
        }
        return max(rest, sell);
    }