# Array - 309. Best Time to Buy and Sell Stock with Cooldown

1. Best Time to Buy and Sell Stock with Cooldown

Say you have an array for which the `i`th 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

1. 定义两个状态
`hold[i]``unhold[i]`分别表示第i天hold股票的最大盈利和不hold股票的最大盈利。
2. 目标是unhold[n-1] (n为交易天数)
3. base case
`hold[0] = -prices[0]`
`hold[1] = max(-prices[1], -prices[0])`
`unhold[0] = 0`
4. 状态转移
hold[i]取最大值：
1、第i天买入 `unhold[i-2] - prices[i]`
2、第i天没有买入 `hold[i-1]`
unhold[i]取最大值：
1、第i天卖出 `hold[i-1] + prices[i]`
2、第i天没有卖出 `unhold[i-1]`

java：

``````class Solution {

public int maxProfit(int[] prices) {
if (prices == null || prices.length == 0) return 0;

int len = prices.length;
int[] hold = new int[len];
int[] unhold = new int[len];

hold[0] = -prices[0];

for (int i = 1; i < len; i++) {
if (i == 1) {
hold[i] = Math.max(hold[i-1], -prices[i]);
} else {
hold[i] = Math.max(hold[i-1], unhold[i-2] - prices[i]);
}

unhold[i] = Math.max(unhold[i-1], hold[i-1] + prices[i]);
}

return unhold[len-1];
}

}
``````

go：

``````func maxProfit(prices []int) int {
if prices == nil || len(prices) == 0 {
return 0
}

var hold = make([]int, len(prices))
var unhold = make([]int, len(prices))

hold[0] = -prices[0]

for i := 1; i < len(prices); i++ {
if i == 1 {
hold[i] = max(-prices[1], hold[0])
} else {
hold[i] = max(unhold[i-2] - prices[i], hold[i-1])
}

unhold[i] = max(hold[i-1] + prices[i], unhold[i-1])
}

return unhold[len(prices) - 1]
}

func max(i, j int) int {
if i > j {
return i
}
return j
}
``````

Nothing just happens, it's all part of a plan.