Day13 股票问题（03）

1、309. 最佳买卖股票时机含冷冻期

• 思路：
  • 买卖有冷冻期，对应一般的股票问题相当于buy[i] = sell[i-2] - price[i]

    注意第1天持有股票的的base case仍为 $Max[dp[0][1],dp[0][0]-price[1]]$

  • coding

    class Solution {
        public int maxProfit(int[] prices) {
            int len = prices.length;
            int[][] dp = new int[len][2];
            for (int i = 0; i < len; i++) {
                if (i == 0) {
                    dp[i][0] = 0;
                    dp[i][1] = -prices[0];
                    continue;
                }
                if (i == 1) {
                    dp[i][0] = Math.max(dp[i - 1][0], dp[i - 1][1] + prices[i]);
                    //第一天持有股票的最大利润 为dp[0][1]  dp[0][0]-prices[i] !!!
                    dp[i][1] = Math.max(dp[i - 1][1], -prices[i]);
                    continue;
                }
                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 - 2][0] - prices[i]);
    
            }
            return dp[len - 1][0];
        }
    }
    

  • 复杂度优化

    class Solution {
        public int maxProfit(int[] prices) {
            int len = prices.length;
            int[] buy = new int[len];
            int[] sell = new int[len];
            for (int i = 0; i < len; i++) {
                if (i == 0) {
                    buy[0] = -prices[i];
                    sell[0] = 0;
                    continue;
                }
                if (i == 1) {
                    //第一天持有股票的最大利润 为dp[0][1]  dp[0][0]-prices[i] !!!
                    buy[i] = Math.max(buy[i - 1], sell[i - 1] - prices[i]);
                    sell[i] = Math.max(sell[i - 1], buy[i - 1] + prices[i]);
                    continue;
                }
                buy[i] = Math.max(buy[i - 1], sell[i - 2] - prices[i]);
                sell[i] = Math.max(sell[i - 1], buy[i - 1] + prices[i]);
            }
            return sell[len - 1];
        }
    }
    class Solution {
        public int maxProfit(int[] prices) {
            int len = prices.length;
            int buy = -prices[0];
            int sell = 0;
            int sellT = 0;
            for (int i = 1; i < len; i++) {
                if (i == 1) {
                    //第一天持有股票的最大利润 为dp[0][1]  dp[0][0]-prices[i] !!!
                    buy = Math.max(buy, sell - prices[i]);
                    sell = Math.max(sell, buy + prices[i]);
                    continue;
                }
                buy = Math.max(buy, sellT - prices[i]);
                sellT = sell;
                sell = Math.max(sell, buy + prices[i]);
    
            }
            return sell;
        }
    }
    

2、714. 买卖股票的最佳时机含手续费

• 思路：
  • 买卖股票有冷冻期，即买的时候价格更高需要扣除手续费
• 题解

   class Solution {
      public int maxProfit(int[] prices, int fee) {
          int n = prices.length;
          int buy = -(prices[0] + fee);
          int sell = 0;
          for (int i = 1; i < n; i++) {
              buy = Math.max(buy, sell - prices[i] - fee);
              sell = Math.max(sell, buy + prices[i]);
          }
          return sell;
      }
  }
  

小结

1. 股票问题主要需要掌握股票的状态转移，即

• $buy[i] = MAX[buy[i-1],sell[i - 1] - price[i]]$
• $sell[i]= MAX[sell[i-1],buy[i - 1] + price[i]]$

2. 注意base case