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前言

一、力扣123.买卖股票的最佳时机III

class Solution {public int maxProfit(int[] prices) {if(prices == null || prices.length == 0){return 0;}int n = prices.length;int[][] dp = new int[n][4];dp[0][0] = -prices[0];dp[0][2] = -prices[0];for(int i = 1; i < n; i ++){dp[i][0] = Math.max(dp[i-1][0], -prices[i]);            //第i天第一次持有dp[i][1] = Math.max(dp[i-1][1], dp[i-1][0] + prices[i]);//第i天第一次不持有dp[i][2] = Math.max(dp[i-1][2], dp[i-1][1] - prices[i]);//第i天第二次持有dp[i][3] = Math.max(dp[i-1][3], dp[i-1][2] + prices[i]);//第i天第二次持有}return dp[n-1][3];}
}

二、力扣● 188.买卖股票的最佳时机IV

class Solution {public int maxProfit(int k, int[] prices) {if(k == 0 || prices == null || prices.length == 0){return 0;}int n = prices.length;int[][] dp = new int[n][2*k+1];for(int j = 1; j < 2*k; j +=2){dp[0][j] = -prices[0];}for(int i = 1; i < n; i ++){for(int j = 0; j < 2*k-1; 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[n-1][2*k];}
}