309.最佳买卖股票时机含冷冻期
class Solution {
public int maxProfit(int[] prices) {
//dp[i][j],第i天状态为j,所剩的最多现金为dp[i][j]
/**
dp[i][0] 持有股票状态
dp[i][1] 保持保持卖出股票状态
dp[i][2] 具体卖出股票状态
dp[i][3] 冷冻期
*/
/**
dp[i][0] = dp[i-1][0]; dp[i-1][3] - price[i]; dp[i-1][1] - prices[i]
dp[i][1] = dp[i-1][1]; dp[i-1][3]
dp[i][2] = dp[i-1][0] + price[i]
dp[i][3] = dp[i-1][2]
*/
int len = prices.length;
if(len == 0) return 0;
int[][] dp = new int[len][4];
dp[0][0] = -prices[0];
for(int i = 1;i<len;i++){
dp[i][0] = Math.max(dp[i-1][0], Math.max(dp[i-1][3] - prices[i],dp[i-1][1] - prices[i]));
dp[i][1] = Math.max(dp[i-1][1],dp[i-1][3]);
dp[i][2] = dp[i-1][0] + prices[i];
dp[i][3] = dp[i-1][2];
}
return Math.max(dp[len-1][1],Math.max(dp[len-1][2],dp[len-1][3]));
}
}
714.买卖股票的最佳时机含手续费
class Solution {
public int maxProfit(int[] prices, int fee) {
//dp[i][j],第i天状态为j,所剩的最多现金为dp[i][j]
/**
dp[i][0] 持有股票状态
dp[i][0] = dp[i-1][0],dp[i-1][1] -price[i]
dp[i][1] 不持有股票状态
dp[i][1] = dp[i-1][1],dp[i-1][0] + price[i] -2
*/
int len = prices.length;
if(len == 0) return 0;
int[][] dp = new int[len][2];
dp[0][0] = -prices[0];
for(int i = 1;i<len;i++){
dp[i][0] = Math.max(dp[i-1][0],dp[i-1][1] -prices[i]);
dp[i][1] = Math.max(dp[i-1][1],dp[i-1][0] + prices[i]-fee);
}
return Math.max(dp[len-1][0],dp[len-1][1]);
}
}