Leetcode 0122. the best time to buy and sell stocks II

【LetMeFly】122. The best time to buy and sell stocks II

Force button topic link ：https://leetcode.cn/problems/best-time-to-buy-and-sell-stock-ii/

Give you an array of integers prices , among  prices[i] Represents a stock i Sky price .

Every day , You can decide whether to buy and / Or sell shares . You at any time   most   Can only hold A wave of Stocks . You can also buy... First , And then in On the same day sell .

return What you can get Maximum profits  .

 

Example 1：

 Input ：prices = [7,1,5,3,6,4]
 Output ：7
 explain ： In the  2  God （ Stock price  = 1） Buy when , In the  3  God （ Stock price  = 5） Sell when ,  The exchange will make a profit  = 5 - 1 = 4 .
      And then , In the  4  God （ Stock price  = 3） Buy when , In the  5  God （ Stock price  = 6） Sell when ,  The exchange will make a profit  = 6 - 3 = 3 .
      The total profit is  4 + 3 = 7 .

Example 2：

 Input ：prices = [1,2,3,4,5]
 Output ：4
 explain ： In the  1  God （ Stock price  = 1） Buy when , In the  5  God  （ Stock price  = 5） Sell when ,  The exchange will make a profit  = 5 - 1 = 4 .
      The total profit is  4 .

Example  3：

 Input ：prices = [7,6,4,3,1]
 Output ：0
 explain ： under these circumstances ,  There is no positive profit from trading , So you can get the maximum profit by not participating in the transaction , The biggest profit is  0 .

 

Tips ：

• 1 <= prices.length <= 3 * 104
• 0 <= prices[i] <= 104

Method 1 ： greedy

In fact, this problem should be simple in medium difficulty .

Since you can buy stocks many times , that As long as I can earn , I'll take it .

Because at most one share is held at the same time , Therefore, in order not to affect my low price purchase behind , As long as you sell it, you can make money , I'll sell .

Then we only need to traverse the array , If tomorrow's stock is more expensive than today , Buy today and sell tomorrow .

That's it .

• Time complexity $O(N)$, among $N$ Is the number of days for which the stock amount is known ($prices.size()$).
• Spatial complexity $O(1)$

AC Code

C++

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

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