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Merge pull request youngyangyang04#2130 from Lozakaka/patch-22

新增java 2*4 solution 並附上對於2-D array, 1-D array的分析

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- 0309.最佳买卖股票时机含冷冻期.md

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`‎problems/0309.最佳买卖股票时机含冷冻期.md‎`

Lines changed: 27 additions & 1 deletion

Original file line number	Diff line number	Diff line change
`@@ -200,7 +200,33 @@ class Solution {`
`200`	`200`	`}`
`201`	`201`	`}`
`202`	`202`	```
`203`		`-`
	`203`	+```java
	`204`	`+//using 2*4 array for space optimization`
	`205`	`+//這裡稍微說一下,我在LeetCode提交的時候,2*4 2-D array的performance基本上和下面的1-D array performance差不多`
	`206`	`+//都是time: 1ms, space: 40.X MB (其實 length*4 的 2-D array也僅僅是space:41.X MB,看起來不多)`
	`207`	`+//股票累的DP題目大致上都是這樣,就當作是一個延伸就好了。真的有人問如何優化,最起碼有東西可以講。`
	`208`	`+class Solution {`
	`209`	`+ /**`
	`210`	`+ 1. [i][0] holding the stock`
	`211`	`+ 2. [i][1] after cooldown but stil not buing the stock`
	`212`	`+ 3. [i][2] selling the stock`
	`213`	`+ 4. [i][3] cooldown`
	`214`	`+ */`
	`215`	`+ public int maxProfit(int[] prices) {`
	`216`	`+ int len = prices.length;`
	`217`	`+ int dp[][] = new int [2][4];`
	`218`	`+ dp[0][0] = -prices[0];`
	`219`	`+`
	`220`	`+ for(int i = 1; i < len; i++){`
	`221`	`+ dp[i % 2][0] = Math.max(Math.max(dp[(i - 1) % 2][0], dp[(i - 1) % 2][1] - prices[i]), dp[(i - 1) % 2][3] - prices[i]);`
	`222`	`+ dp[i % 2][1] = Math.max(dp[(i - 1) % 2][1], dp[(i - 1) % 2][3]);`
	`223`	`+ dp[i % 2][2] = dp[(i - 1) % 2][0] + prices[i];`
	`224`	`+ dp[i % 2][3] = dp[(i - 1) % 2][2];`
	`225`	`+ }`
	`226`	`+ return Math.max(Math.max(dp[(len - 1) % 2][1], dp[(len - 1) % 2][2]), dp[(len - 1) % 2][3]);`
	`227`	`+ }`
	`228`	`+}`
	`229`	+```
`204`	`230`	```java
`205`	`231`	`// 一维数组优化`
`206`	`232`	`class Solution {`

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