题目

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:Input: [1,2,3,0,2]
Output: 3
Explanation: transactions = [buy, sell, cooldown, buy, sell]

解题思路:
第一种方案, 假设数组长度为n, dp[i][j]为从i到j所能达到的最大收益，那么本题即求dp[0][n - 1],
对于dp[i][j], 其可能的cooldown位置有 I, i + 1, ..., j - 1, j,
所以存在递推关系
dp[i][j] = max{ dp[i][k - 1] + dp[k + 1][j]} k = i, i + 1, ... , j - 1, j
当k == i 时, dp[i][k - 1] 不存在，即只有dp[k + 1][j], 同理
当k == j 时, dp[k + 1][j] 不存在，即只有dp[i][k - 1]
prices[j] - prices[I] 为dp[I][j]的初始值
所以最终dp[i][j] = max(prices[j] - prices[I], max{dp[i][k - 1] + dp[k + 1][j]})
而题目希望求解的是dp[0][n - 1]. 所以i 从n-1往0求解，j从0往n-1求解
时间复杂度O(n^3) 空间复杂度O(n^2)

代码如下

class Solution {
public://suppose cooldown at k//dp[i][j] = max{dp[i][k - 1] + dp[k + 1][j]} k = i ... jint maxProfit(vector<int>& prices) {int n = prices.size();if (0 == n) return 0;vector<vector<int>> dp(n, vector<int>(n, 0));for (int i = 0; i < n; i++) {for (int j = i + 1; j < n; j++) {dp[i][j] = prices[j] - prices[i];}}for (int i = n - 1; i >= 0; i--) {for (int j = 0; j < n; j++) {//cout<<"i="<<i<<" j="<<j<<" "<<dp[i][j]<<endl;for (int k = i; k < j; k++) {int tmp = 0;if (k - 1 >= i) {tmp += dp[i][k - 1];}if (k + 1 <= j) {tmp += dp[k + 1][j];}dp[i][j] = max(dp[i][j], tmp);}}}return dp[0][n - 1];}
};

第二种方案：顺序DP
常规的DP的类型主要有三类，矩阵dp，一个一维数组的dp，两个一维数组的dp
矩阵dp 构造f[i][j], 一维dp构造f[i], 两个一维dp构造f[i][j]
本题恰好可以使用顺序dp，而且是一维的数组

解题思路：
每一天股票的持有状态可能有三种情况
cool down-->buy-->sell-->cool down-->buy-->sell-->cool down
状态转换的关系如上， leetcode讨论区有人画了状态图，非常容易理解, 参考链接
https://leetcode.com/explore/interview/card/top-interview-questions-hard/121/dynamic-programming/862/discuss/75928/Share-my-DP-solution-(By-State-Machine-Thinking)

也就是说
buy的状态 可能是从前一个buy 或者前一个cool down过来
sell的状态 只能是从前一个buy过来
cool down的状态 可能是从前一个cool down或者前一个sell的状态过来

这里需要搞清楚
1)sell 和 cool down的区别， sell状态只有 卖出的那个时刻状态是保持的， 卖完第二天状态就是cool down了.
2)buy 到 sell 之间的这段时间，按题意并不算cool down，而全是buy状态
3)sell 到 cool down之间的这段时间，全是cool down状态

由此可以得出
buy[i] = max(buy[i - 1], rest[i - 1] - prices[I]) // 这里用rest 表示 cool down
rest[i] = max(rest[i - 1], sell[I - 1])
sell[I] = buy[I - 1] + prices[i]

代码如下
java

class Solution {public int maxProfit(int[] prices) {int n = prices.length;if (0 == n) return 0;int[] buy = new int[n];int[] rest = new int[n];int[] sell = new int[n];buy[0] = -prices[0];rest[0] = 0;sell[0] = Integer.MIN_VALUE;for (int i = 1; i < n; i++) {buy[i] = Math.max(buy[i - 1], rest[i - 1] - prices[i]);rest[i] = Math.max(rest[i - 1], sell[i - 1]);sell[i] = buy[i - 1] + prices[i];}return Math.max(rest[n - 1], sell[n - 1]);}
}

c++

class Solution {
public:int maxProfit(vector<int>& prices) {int n = prices.size();if (0 == n) return 0;vector<int> buy(n, 0);vector<int> rest(n, 0);vector<int> sell(n, 0);buy[0] = -prices[0];rest[0] = 0;//不可能存在，所以收益取最小，因为i位置，我们希望取的是最大值，//将sell设置为最小值，表示永远不可能取该值sell[0] = INT_MIN;for (int i = 1; i < n; i++) {buy[i] = max(buy[i - 1], rest[i - 1] - prices[i]);rest[i] = max(rest[i - 1], sell[i - 1]);sell[i] = buy[i - 1] + prices[i];}return max(rest[n - 1], sell[n - 1]);}
};