leetcode每日一题买卖股票的最佳时机III

    public int maxProfit(int[] prices) {
        long[][] dp = new long[prices.length + 1][5];
        for (long[] dayStatus : dp) {
            Arrays.fill(dayStatus, Integer.MIN_VALUE);
        }
        dp[0][0] = 0; // 第零天的“没买过”状态的收入值是 0，其他都为 Integer.MIN_VALUE 表示没计算过
        for (int i = 1; i < dp.length; i++) {
            dp[i][0] = 0;
            dp[i][1] = Math.max(dp[i - 1][1], dp[i - 1][0] - prices[i - 1]);
            dp[i][2] = Math.max(dp[i - 1][2], dp[i - 1][1] + prices[i - 1]);
            dp[i][3] = Math.max(dp[i - 1][3], dp[i - 1][2] - prices[i - 1]);
            dp[i][4] = Math.max(dp[i - 1][4], dp[i - 1][3] + prices[i - 1]);
        }
        int res = Integer.MIN_VALUE;
        for (int i = 0; i < dp[prices.length].length; i++) {
            res = Math.max(res, (int)dp[prices.length][i]);
        }
        return res;
    }