DAY42股票问题II

IV

可以k次买卖

    public int maxProfit(int k, int[] prices) {
        int[][]dp=new int[prices.length][2*k+1];
        //dp[i][2*j+1]第i天第j次买卖
        for(int i=0;i<k;i++){
            dp[0][i*2+1]=-prices[0];
        }
        for(int i=1;i<prices.length;i++){
            for(int j=0;j<k*2-1;j=j+2){
                dp[i][j+1]=Math.max(dp[i-1][j+1],dp[i-1][j]-prices[i]);
                dp[i][j+2]=Math.max(dp[i-1][j+2],dp[i-1][j+1]+prices[i]);

            }
        }
        return dp[prices.length-1][k*2];


    }

含冷冻期

    public int maxProfit(int[] prices) {
        int[][]dp=new int[prices.length][3];
        //0不持有且当天没卖出
        //1持有
        //2不持有且当天卖出了
        dp[0][0]=0;
        dp[0][1]=-prices;
        dp[0][2]=0;
        for(int i=1;i<prices.length;i++){
            dp[i][0]=Math.max(dp[i-1][0],dp[i-1][2]);
            dp[i][1]=Math.max(dp[i-1][1],dp[i-1][0]-prices[i]);
            dp[i][2]=dp[i-1][1]+prices[i];
        }
        return Math.max(dp[n-1][0],dp[n-1][2]);


    }

含手续费

    public int maxProfit(int[] prices, int fee) {
        int [][]dp=new int[prices.length][2];
        dp[0][0]=-prices[0];
         for (int i = 1; i < prices.length; i++) {
        dp[i][0] = Math.max(dp[i - 1][0], dp[i - 1][1] - prices[i]);
        dp[i][1] = Math.max(dp[i - 1][0] + prices[i] - fee, dp[i - 1][1]);
    }
        return Math.max(dp[prices.length - 1][0], dp[prices.length - 1][1]);

    }