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]);}