Dynamic planning for solving problems (5)

21. The best time to buy and sell stocks

   class Solution {
    
    public:
        int maxProfit(vector<int>& prices) {
    
            int len = prices.size();
            if (len == 0) return 0;
            vector<vector<int>> dp(len, vector<int>(2));
            dp[0][0] -= prices[0];
            dp[0][1] = 0;
            for (int i = 1; i < len; i++) {
    
                dp[i][0] = max(dp[i - 1][0], -prices[i]);
                dp[i][1] = max(dp[i - 1][1], prices[i] + dp[i - 1][0]);
            }
            return dp[len - 1][1];
        }
    };

122. The best time to buy and sell stocks II

  class Solution {
    
    public:
        int maxProfit(vector<int>& prices) {
    
            int len = prices.size();
            vector<vector<int>> dp(len, vector<int>(2, 0));
            dp[0][0] -= prices[0];
            dp[0][1] = 0;
            for (int i = 1; i < len; i++) {
    
                dp[i][0] = max(dp[i - 1][0], dp[i - 1][1] - prices[i]); //  Notice here is and 121.  The only difference between the best time to buy and sell stocks .
                dp[i][1] = max(dp[i - 1][1], dp[i - 1][0] + prices[i]);
            }
            return dp[len - 1][1];
        }
    };

• Time complexity ：O(n)
• Spatial complexity ：O(n)

123. The best time to buy and sell stocks III

dpi in i It means the first one i God ,j by [0 - 4] Five states ,dpi It means the first one i Day state j The biggest cash left .

//  Version of a 
class Solution {
    
public:
    int maxProfit(vector<int>& prices) {
    
        if (prices.size() == 0) return 0;
        vector<vector<int>> dp(prices.size(), vector<int>(5, 0));
        dp[0][1] = -prices[0];
        dp[0][3] = -prices[0];
        for (int i = 1; i < prices.size(); i++) {
    
            dp[i][0] = dp[i - 1][0];
            dp[i][1] = max(dp[i - 1][1], dp[i - 1][0] - prices[i]);
            dp[i][2] = max(dp[i - 1][2], dp[i - 1][1] + prices[i]);
            dp[i][3] = max(dp[i - 1][3], dp[i - 1][2] - prices[i]);
            dp[i][4] = max(dp[i - 1][4], dp[i - 1][3] + prices[i]);
        }
        return dp[prices.size() - 1][4];
    }
};

188. The best time to buy and sell stocks IV

So in the same way, we can deduce dp0 When j When it is an odd number, it is initialized to -prices[0]

initialization All worth , The biggest difference is the analogy here j Odd is to buy , Even numbers are sold .

  class Solution {
    
    public:
        int maxProfit(int k, vector<int>& prices) {
    
    
            if (prices.size() == 0) return 0;
            vector<vector<int>> dp(prices.size(), vector<int>(2 * k + 1, 0));
            for (int j = 1; j < 2 * k; j += 2) {
    
                dp[0][j] = -prices[0];
            }
            for (int i = 1;i < prices.size(); i++) {
    
                for (int j = 0; j < 2 * k - 1; j += 2) {
    
                    dp[i][j + 1] = max(dp[i - 1][j + 1], dp[i - 1][j] - prices[i]);
                    dp[i][j + 2] = max(dp[i - 1][j + 2], dp[i - 1][j + 1] + prices[i]);
                }
            }
            return dp[prices.size() - 1][2 * k];
        }
    };

309. The best time to buy and sell stocks includes the freezing period

summary ： existing 2 A buy sell Status of holding shares , Only then Buy today Selling status today .

• State one ： Buy stock status （ Buy stocks today , Or you bought stocks before and didn't operate ）
• Status of selling shares , There are two states of selling stocks
  • State two ： Sold the stock two days ago , Passed the freezing period , Never operated , Keep selling stocks today
  • State three ： I sold shares today
• State four ： Today is in a frozen state , But the frozen state is unsustainable , Only one day ！

Incomprehensible state ,

• State one ： Buy stock status （ Buy stocks today , Or you bought stocks before and didn't operate ）
• Status of selling shares , There are two states of selling stocks
  • State two ： Sold the stock two days ago , Passed the freezing period , Never operated , Keep selling stocks today
  • State three ： I sold shares today
• State four ： Today is in a frozen state , But the frozen state is unsustainable , Only one day ！

Write the wrong code ：

 dp[i][0] = max(dp[i - 1][0], max(dp[i - 1][3], dp[i - 1][1]) - prices[i]); //max(dp[i - 2][3]
 max(dp[i - 1][3], dp[i - 1][1]) - prices[i]);

class Solution {
    
public:
    int maxProfit(vector<int>& prices) {
    
        int n = prices.size();
        if (n == 0) return 0;
        vector<vector<int>> dp(n, vector<int>(4, 0));
        dp[0][0] -= prices[0]; //  Holding shares 
        for (int i = 1; i < n; i++) {
    
            dp[i][0] = max(dp[i - 1][0], max(dp[i - 1][3], dp[i - 1][1]) - prices[i]);
            dp[i][1] = 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 max(dp[n - 1][3],max(dp[n - 1][1], dp[n - 1][2]));
    }
};

714. The best time to buy and sell stocks includes handling fees

The difference is that there needs to be an additional operation to deduct the handling fee .

  class Solution {
    
    public:
        int maxProfit(vector<int>& prices, int fee) {
    
            int n = prices.size();
            vector<vector<int>> dp(n, vector<int>(2, 0));
            dp[0][0] -= prices[0]; //  Holding shares 
            for (int i = 1; i < n; i++) {
    
                dp[i][0] = max(dp[i - 1][0], dp[i - 1][1] - prices[i]);
                dp[i][1] = max(dp[i - 1][1], dp[i - 1][0] + prices[i] - fee);
            }
            return max(dp[n - 1][0], dp[n - 1][1]);
        }
    };

• Time complexity ：O(n)
• Spatial complexity ：O(n)