123.买卖股票的最佳时机III

func maxProfit(prices []int) int {dp := make([][]int , len(prices))dp[0] = []int{0, -prices[0], 0, -prices[0], 0}for i := 1; i < len(prices);i++{val0 := dp[i - 1][0]val1 := max(dp[i - 1][0] - prices[i], dp[i - 1][1])val2 := max(dp[i - 1][1] + prices[i], dp[i - 1][2])val3 := max(dp[i - 1][2] - prices[i], dp[i - 1][3])val4 := max(dp[i - 1][3] + prices[i], dp[i - 1][4])dp[i] = []int{val0, val1, val2, val3, val4}}return dp[len(prices)- 1][4]
}
func max(a, b int)int{if a < b{return b}return a
}

● 188.买卖股票的最佳时机IV
和买卖股票3中的思路一样，只不过从两次换成了k次

func maxProfit(k int, prices []int) int {dp := make([][]int, len(prices))for i := 0; i < len(dp); i++{tmp := make([]int, 2 * k + 1)dp[i] = tmpif i == 0{for j := 1; j < 2 * k + 1; j += 2{dp[i][j] = -prices[0]}}}for i := 1; i < len(dp); i++{dp[i][0] = dp[i - 1][0]for j :=1; j < 2 * k + 1; j += 2{dp[i][j] = max(dp[i - 1][j - 1] - prices[i], dp[i - 1][j])dp[i][j + 1] = max(dp[i - 1][j] + prices[i], dp[i - 1][j + 1])}}return dp[len(prices) - 1][2 * k]
}
func max(a , b int)int{if a < b{return b}return a
}