The best time to buy and sell stocks (leetcode-121)

The best time to buy and sell stocks （LeetCode-121）

subject

Given an array prices , It's the first i Elements prices[i] Represents the number of shares in a given stock i Sky price .

You can only choose One day Buy this stock , And choose A different day in the future Sell the stock . Design an algorithm to calculate the maximum profit you can get .

Return the maximum profit you can make from the deal . If you can't make any profit , return 0 .

Example 1：

 Input ：[7,1,5,3,6,4]
 Output ：5
 explain ： In the  2  God （ Stock price  = 1） Buy when , In the  5  God （ Stock price  = 6） Sell when , Maximum profit  = 6-1 = 5 .
      Note that profit cannot be  7-1 = 6,  Because the selling price needs to be higher than the buying price ; meanwhile , You can't sell stocks before you buy them .

Example 2：

 Input ：prices = [7,6,4,3,1]
 Output ：0
 explain ： under these circumstances ,  No deal is done ,  So the biggest profit is  0.

Tips ：

• 1 <= prices.length <= 105
• 0 <= prices[i] <= 104

Ideas

1. dp[i][ take 0 or 1] The meaning of

   • $dp[i][0]$ It means the first one $i$ Cash from holding the stock for days
   • $dp[i][1]$ It means the first one $i$ Cash from not holding the stock for days

2. The recursive formula

   • $dp[i][0]$ Can be derived from two states
     • The first $i-1$ Days holding shares , It is equal to $dp[i-1][0]$
     • The first $i$ Days to buy shares , The cash will be after buying today $-price[i]$
     • Choose the one with the most cash , That is, the greater of the two
   • $dp[i][1]$ Can be derived from two states
     • The first $i-1$ Days do not hold shares , It is equal to $dp[i-1][0]$
     • The first $i$ Days to sell shares , It is equal to $price[i]+dp[i-1][0]$
     • Choose the one with the most cash , That is, the greater of the two

3. Array initialization

   • dp[0][0] It means the first one 0 Days holding shares , So it's equal to $-price[0]$
   • dp[0][1] It means the first one 0 Days do not hold shares , be equal to 0

4. traversal order

   • Before and after

5. The test case

Go through the use case in your own mind and you'll understand

Code display

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

Rolling array optimization ！

As can be seen from the recurrence formula ,dp[i] Just relying on dp[i - 1] The state of .

So just record the status of the previous day and the current day

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

This optimization method is really to understand the binary , I was shocked .