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买卖股票的最佳时机

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class Solution {
    public int maxProfit(int[] prices) {
        int n = prices.length;
        int dp_i_0 = 0, dp_i_1 = Integer.MIN_VALUE;
        for (int i = 0; i < n; i++) {
            dp_i_0 = Math.max(dp_i_0, dp_i_1 + prices[i]);
            dp_i_1 = Math.max(dp_i_1, -prices[i]);
        }
        return dp_i_0;
    }
}

买卖股票的最佳时机 II

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class Solution {
    public int maxProfit(int[] prices) {
        int n = prices.length;
        int dp_i_0 = 0, dp_i_1 = Integer.MIN_VALUE;
        for (int i = 0; i < n; i++) {
            int temp = dp_i_0;
            dp_i_0 = Math.max(dp_i_0, dp_i_1 + prices[i]);
            dp_i_1 = Math.max(dp_i_1, temp - prices[i]);
        }
        return dp_i_0;
    }
}

最佳买卖股票时机含冷冻期

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class Solution {
    public int maxProfit(int[] prices) {
        int n = prices.length;
        int dp_i_0 = 0, dp_i_1 = Integer.MIN_VALUE;
        int dp_pre_0 = 0;
        for (int i = 0; i < n; i++) {
            int temp = dp_i_0;
            dp_i_0 = Math.max(dp_i_0, dp_i_1 + prices[i]);
            dp_i_1 = Math.max(dp_i_1, dp_pre_0 - prices[i]);
            dp_pre_0 = temp;
        }
        return dp_i_0;
    }
}

买卖股票的最佳时机含手续费

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class Solution {
    public int maxProfit(int[] prices, int fee) {
        int n = prices.length;
        int dp_i_0 = 0, dp_i_1 = Integer.MIN_VALUE;
        for (int i = 0; i < n; i++) {
            int temp = dp_i_0;
            dp_i_0 = Math.max(dp_i_0, dp_i_1 + prices[i]);
            dp_i_1 = Math.max(dp_i_1, temp - prices[i] - fee);
        }
        return dp_i_0;
    }
}

买卖股票的最佳时机 III

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class Solution {
    public int maxProfit(int[] prices) {
        int dp_i10 = 0, dp_i11 = Integer.MIN_VALUE;
        int dp_i20 = 0, dp_i21 = Integer.MIN_VALUE;
        for (int i = 0; i < prices.length; i++) {
            dp_i20 = Math.max(dp_i20, dp_i21 + prices[i]);
            dp_i21 = Math.max(dp_i21, dp_i10 - prices[i]);
            dp_i10 = Math.max(dp_i10, dp_i11 + prices[i]);
            //第一次购买
            dp_i11 = Math.max(dp_i11, -prices[i]);
        }
        return dp_i20;
    }
}

买卖股票的最佳时机 IV

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class Solution {
    public int maxProfit(int k, int[] prices) {
        int n = prices.length;
        if (k > n / 2) {
            return maxProfit_k_inf(prices);
        }
        int[][][] dp = new int[n][k + 1][2];
        for (int i = 0; i < n; i++) {
            for (int j = k; j >= 1; j--) {
                if (i == 0) {
                    //Base Case
                    dp[i][j][0] = 0;
                    dp[i][j][1] = -prices[i];
                }else {
                    //注意i和k的变化
                    dp[i][j][0] = Math.max(dp[i - 1][j][0], dp[i - 1][j][1] + prices[i]);
                    dp[i][j][1] = Math.max(dp[i - 1][j][1], dp[i - 1][j - 1][0] - prices[i]);
                }
            }
        }
        return dp[n - 1][k][0];
    }

    private int maxProfit_k_inf(int[] prices) {
        int n = prices.length;
        int dp_i_0 = 0, dp_i_1 = Integer.MIN_VALUE;
        for (int i = 0; i < n; i++) {
            int temp = dp_i_0;
            dp_i_0 = Math.max(dp_i_0, dp_i_1 + prices[i]);
            dp_i_1 = Math.max(dp_i_1, temp - prices[i]);
        }
        return dp_i_0;
    }
}

总结

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dp[i][k][0 or 1]
i为天数，k为交易次数，0 1表示当前是否持有股票
0 <= i <= n-1, 1 <= k <= K
n 为天数，大 K 为最多交易数
此问题共 n × K × 2 种状态，全部穷举就能搞定。

for 0 <= i < n:
    for 1 <= k <= K:
        for s in {0, 1}:
            dp[i][k][s] = max(buy, sell, rest)


base case：
dp[-1][k][0] = dp[i][0][0] = 0
dp[-1][k][1] = dp[i][0][1] = -infinity

状态转移方程：
dp[i][k][0] = max(dp[i-1][k][0], dp[i-1][k][1] + prices[i])
dp[i][k][1] = max(dp[i-1][k][1], dp[i-1][k-1][0] - prices[i])