188. 买卖股票的最佳时机 IV - 力扣（LeetCode）

class Solution {public int maxProfit(int k, int[] prices) {int[][] dp = new int[prices.length][2*k];for(int i=0;i<2*k;i++){if(i%2 == 0){dp[0][i] = -prices[0];}else{dp[0][i] = 0;}    }for(int i=1;i<prices.length;i++){for(int j=0;j<2*k;j++){if(j%2==0){ //持有if(j==0){dp[i][j] = Math.max(dp[i-1][j],-prices[i]);}else{dp[i][j] = Math.max(dp[i-1][j],-prices[i] + dp[i-1][j-1]);}}else{      //不持有dp[i][j] = Math.max(dp[i-1][j],prices[i] + dp[i-1][j-1]);}}}int result = 0;for(int i=0;i<2*k;i++){if(i%2 != 0){result = Math.max(result,dp[prices.length-1][i]);}  }return result;}
}

309. 买卖股票的最佳时机含冷冻期 - 力扣（LeetCode） 

class Solution {public int maxProfit(int[] prices) {if(prices.length <= 1){return 0;}int[][] dp = new int[prices.length][4];dp[0][0] = -prices[0]; //第i天持有状态的最大收益dp[0][1] = 0;          //第i天保持卖出状态的收益dp[0][2] = 0;          //第i天卖出dp[0][3] = 0;          //第i天为冻结器的收益for(int i=1;i<prices.length;i++){dp[i][0] = Math.max(Math.max(dp[i-1][0],-prices[i]+dp[i-1][1]),-prices[i]+dp[i-1][3]);dp[i][1] = Math.max(dp[i-1][1],dp[i-1][3]);dp[i][2] = prices[i] + dp[i-1][0];dp[i][3] = dp[i-1][2];} return Math.max(Math.max(dp[prices.length-1][1],dp[prices.length-1][2]),dp[prices.length-1][3]);}
}

714. 买卖股票的最佳时机含手续费 - 力扣（LeetCode） 

class Solution {public int maxProfit(int[] prices, int fee) {int[][] dp = new int[prices.length][2];dp[0][0] = -prices[0]; //第i天持有dp[0][1] = 0;          //第i天不持有for(int i=1;i<prices.length;i++){dp[i][0] = Math.max(dp[i-1][0],-prices[i]+dp[i-1][1]);dp[i][1] = Math.max(dp[i-1][1],prices[i]+dp[i-1][0]-fee);}return dp[prices.length-1][1];}
}