Merge pull request #896 from RyouMon/master

新增最佳买卖股票时机系列题目的GO解法

problems/0122.买卖股票的最佳时机II（动态规划）.md

@@ -196,6 +196,33 @@ class Solution:
 ```
 
 Go：
+```go
+// 买卖股票的最佳时机Ⅱ 动态规划
+// 时间复杂度O(n) 空间复杂度O(n)
+func maxProfit(prices []int) int {
+    dp := make([][]int, len(prices))
+    status := make([]int, len(prices) * 2)
+    for i := range dp {
+        dp[i] = status[:2]
+        status = status[2:]
+    }
+    dp[0][0] = -prices[0]
+    
+    for i := 1; i < len(prices); i++ {
+        dp[i][0] = max(dp[i - 1][0], dp[i - 1][1] - prices[i])
+        dp[i][1] = max(dp[i - 1][1], dp[i - 1][0] + prices[i])
+    }
+    
+    return dp[len(prices) - 1][1]
+}
+
+func max(a, b int) int {
+    if a > b {
+        return a
+    }
+    return b
+}
+```
 
 ```go
 func maxProfit(prices []int) int {

problems/0123.买卖股票的最佳时机III.md

@@ -346,7 +346,38 @@ const maxProfit = prices => {
 };
 ```
 
+Go:
 
+> 版本一：
+```go
+// 买卖股票的最佳时机III 动态规划
+// 时间复杂度O(n) 空间复杂度O(n)
+func maxProfit(prices []int) int {
+    dp := make([][]int, len(prices))
+    status := make([]int, len(prices) * 4)
+    for i := range dp {
+        dp[i] = status[:4]
+        status = status[4:]
+    }
+    dp[0][0], dp[0][2] = -prices[0], -prices[0]
+    
+    for i := 1; i < len(prices); i++ {
+        dp[i][0] = max(dp[i - 1][0], -prices[i])
+        dp[i][1] = max(dp[i - 1][1], dp[i - 1][0] + prices[i])
+        dp[i][2] = max(dp[i - 1][2], dp[i - 1][1] - prices[i])
+        dp[i][3] = max(dp[i - 1][3], dp[i - 1][2] + prices[i])
+    }
+    
+    return dp[len(prices) - 1][3]
+}
+
+func max(a, b int) int {
+    if a > b {
+        return a
+    }
+    return b
+}
+```
 
 
 -----------------------

problems/0188.买卖股票的最佳时机IV.md

@@ -272,6 +272,41 @@ class Solution:
         return dp[2*k]
 ```
 Go：
+版本一：
+```go
+// 买卖股票的最佳时机IV 动态规划
+// 时间复杂度O(kn) 空间复杂度O(kn)
+func maxProfit(k int, prices []int) int {
+    if k == 0 || len(prices) == 0 {
+        return 0
+    }
+    
+    dp := make([][]int, len(prices))
+    status := make([]int, (2 * k + 1) * len(prices))
+    for i := range dp {
+        dp[i] = status[:2 * k + 1]
+        status = status[2 * k + 1:]
+    }
+    for j := 1; j < 2 * k; j += 2 {
+        dp[0][j] = -prices[0]
+    }
+    
+    for i := 1; i < len(prices); i++ {
+        for j := 0; j < 2 * k; j += 2 {
+            dp[i][j + 1] = max(dp[i - 1][j + 1], dp[i - 1][j] - prices[i])
+            dp[i][j + 2] = max(dp[i - 1][j + 2], dp[i - 1][j + 1] + prices[i])
+        }
+    }
+    return dp[len(prices) - 1][2 * k]
+}
+
+func max(a, b int) int {
+    if a > b {
+        return a
+    }
+    return b
+}
+```
 
 ```go
 func maxProfit(k int, prices []int) int {

problems/0309.最佳买卖股票时机含冷冻期.md

@@ -204,6 +204,40 @@ class Solution:
 ```
 
 Go：
+```go
+// 最佳买卖股票时机含冷冻期 动态规划
+// 时间复杂度O(n) 空间复杂度O(n)
+func maxProfit(prices []int) int {
+    n := len(prices)
+    if n < 2 {
+        return 0
+    }
+
+    dp := make([][]int, n)
+    status := make([]int, n * 4)
+    for i := range dp {
+        dp[i] = status[:4]
+        status = status[4:]
+    }
+    dp[0][0] = -prices[0]
+    
+    for i := 1; i < n; i++ {
+        dp[i][0] = max(dp[i - 1][0], max(dp[i - 1][1] - prices[i], dp[i - 1][3] - prices[i]))
+        dp[i][1] = max(dp[i - 1][1], dp[i - 1][3])
+        dp[i][2] = dp[i - 1][0] + prices[i]
+        dp[i][3] = dp[i - 1][2]
+    }
+    
+    return max(dp[n - 1][1], max(dp[n - 1][2], dp[n - 1][3]))
+}
+
+func max(a, b int) int {
+    if a > b {
+        return a
+    }
+    return b
+}
+```
 
 Javascript:
 

problems/0714.买卖股票的最佳时机含手续费（动态规划）.md

@@ -149,14 +149,16 @@ class Solution:
 ```
 
 Go：
-```Go
+```go
+// 买卖股票的最佳时机含手续费 动态规划
+// 时间复杂度O(n) 空间复杂度O(n)
 func maxProfit(prices []int, fee int) int {
     n := len(prices)
     dp := make([][2]int, n)
     dp[0][0] = -prices[0]
     for i := 1; i < n; i++ {
-        dp[i][1] = max(dp[i-1][1], dp[i-1][0]+prices[i]-fee)
+        dp[i][1] = max(dp[i-1][1], dp[i-1][0] + prices[i] - fee)
-        dp[i][0] = max(dp[i-1][0], dp[i-1][1]-prices[i])
+        dp[i][0] = max(dp[i-1][0], dp[i-1][1] - prices[i])
     }
     return dp[n-1][1]
 }