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更正了188股票问题一维数组空间优化的java代码

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problems
- 0188.买卖股票的最佳时机IV.md

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`‎problems/0188.买卖股票的最佳时机IV.md`

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Original file line number	Diff line number	Diff line change
`@@ -165,7 +165,7 @@ public:`
`165`	`165`
`166`	`166`	`## 其他语言版本`
`167`	`167`
`168`		`-Java:`
	`168`	`+### Java`
`169`	`169`
`170`	`170`	```java
`171`	`171`	`// 版本一: 三维 dp数组`
`@@ -228,9 +228,9 @@ class Solution {`
`228`	`228`	`if(k == 0){`
`229`	`229`	`return 0;`
`230`	`230`	`}`
`231`		`- // 其实就是123题的扩展,123题只用记录2天的状态`
`232`		`- // 这里记录k天的状态就行了`
`233`		`- // 每天都有买入,卖出两个状态,所以要乘 2`
	`231`	`+ // 其实就是123题的扩展,123题只用记录2次交易的状态`
	`232`	`+ // 这里记录k次交易的状态就行了`
	`233`	`+ // 每次交易都有买入,卖出两个状态,所以要乘 2`
`234`	`234`	`int[] dp = new int[2 * k];`
`235`	`235`	`// 按123题解题格式那样,做一个初始化`
`236`	`236`	`for(int i = 0; i < dp.length / 2; i++){`
`@@ -246,15 +246,15 @@ class Solution {`
`246`	`246`	`dp[j + 1] = Math.max(dp[j + 1], dp[j] + prices[i - 1]);`
`247`	`247`	`}`
`248`	`248`	`}`
`249`		`- // 返回最后一天卖出状态的结果就行了`
	`249`	`+ // 返回最后一次交易卖出状态的结果就行了`
`250`	`250`	`return dp[dp.length - 1];`
`251`	`251`	`}`
`252`	`252`	`}`
`253`	`253`	```
`254`	`254`
`255`		`-`
`256`		`-Python:`
	`255`	`+### Python`
`257`	`256`	`版本一`
	`257`	`+`
`258`	`258`	```python
`259`	`259`	`class Solution:`
`260`	`260`	`def maxProfit(self, k: int, prices: List[int]) -> int:`
`@@ -285,8 +285,9 @@ class Solution:`
`285`	`285`	`dp[j] = max(dp[j],dp[j-1]+prices[i])`
`286`	`286`	`return dp[2*k]`
`287`	`287`	```
`288`		`-Go:`
	`288`	`+### Go`
`289`	`289`	`版本一:`
	`290`	`+`
`290`	`291`	```go
`291`	`292`	`// 买卖股票的最佳时机IV 动态规划`
`292`	`293`	`// 时间复杂度O(kn) 空间复杂度O(kn)`
`@@ -356,10 +357,7 @@ func max(a,b int)int{`
`356`	`357`	`}`
`357`	`358`	```
`358`	`359`
`359`		`-`
`360`		`-`
`361`		`-`
`362`		`-Javascript:`
	`360`	`+### Javascript`
`363`	`361`
`364`	`362`	```javascript
`365`	`363`	`// 方法一:动态规划`

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