*** Leetcode 309. Best Time to Buy and Sell Stock with Cooldown | dp

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

因为做过之前那个题，所以还是能比较快做出来的。

做法1：

dp[i][0]: 第i天卖，完成i次买和卖

dp[i][1]:第i天买，完成i-1次卖和i次买

class Solution {
public:int maxProfit(vector<int>& prices) {if (!prices.size()) return 0;// vector <int> dp(prices.size(), 0);int dp[prices.size()][2];int ans = 0;for (int i = 0; i < prices.size(); i++) {dp[i][0] = 0;dp[i][1] = INT_MIN;}for (int i = 0; i < prices.size(); i++) {if (i <= 1) {dp[i][1] = -prices[i];}for (int j = 0; j < i; j++) {if (j < i - 1)dp[i][1] = max(dp[i][1], dp[j][0] - prices[i]);dp[i][0] = max(dp[i][0], dp[j][1] + prices[i]);}ans = max(ans, dp[i][0]);// cout << "i:" << i << " d0:" << dp[i][0] << " d1:" << dp[i][1] << endl;}return ans;}
};

更自然的解法：https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-cooldown/discuss/75928/Share-my-DP-solution-(By-State-Machine-Thinking)

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

从这题和之前的卖股票的那个题，学到了状态转移表示dp的思路

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