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新增 0188.买卖股票的最佳时机IV.md Go解法
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@ -275,6 +275,41 @@ class Solution:
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return dp[2*k]
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```
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Go:
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版本一:
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```go
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// 买卖股票的最佳时机IV 动态规划
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// 时间复杂度O(kn) 空间复杂度O(kn)
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func maxProfit(k int, prices []int) int {
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if k == 0 || len(prices) == 0 {
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return 0
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}
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dp := make([][]int, len(prices))
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status := make([]int, (2 * k + 1) * len(prices))
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for i := range dp {
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dp[i] = status[:2 * k + 1]
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status = status[2 * k + 1:]
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}
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for j := 1; j < 2 * k; j += 2 {
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dp[0][j] = -prices[0]
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}
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for i := 1; i < len(prices); i++ {
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for j := 0; j < 2 * k; j += 2 {
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dp[i][j + 1] = max(dp[i - 1][j + 1], dp[i - 1][j] - prices[i])
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dp[i][j + 2] = max(dp[i - 1][j + 2], dp[i - 1][j + 1] + prices[i])
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}
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}
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return dp[len(prices) - 1][2 * k]
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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Javascript:
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