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The best time to stock trading
The best time to stock trading
Given an array, which i-th element is a given stock i-day prices.
If you are only allowed up to complete a transaction (ie buying and selling a stock), to design an algorithm to compute the maximum profit you can get.
Note that you can not sell the stock before buying stocks.
Example 1:
Input: [7,1,5,3,6,4]
Output: 5
Explanation: Day 2 (= 1 stock price), when buying, on day 5 (stock price = 6), when sold, The maximum profit = 6-1 = 5.
Note that profits can not be 7-1 = 6, because the selling price must be greater than the purchase price. \
Example 2:
Enter: [7,6,4,3,1]
Output: 0
to explain: In this case, no transaction is completed, the maximum profit is zero.
Thinking
Violence 1.
[7,1,5,3,6,4]
Take 7, a view from the end to all numbers greater than 7, the difference operator, taking the maximum
take a view from the end of all greater than 5 to 1 number, count the difference, taking the maximum
...
int maxProfit(vector<int>& prices) {
if (prices.size() < 2)
{
return 0;
}
int i = 0;
int maxPrice = 0;
for (;i < prices.size(); i++)
{
int endPos = prices.size() - 1;
for(int j = endPos; j > i; j--)
{
int p = prices[j] - prices[i];
if(p > 0 && p > maxPrice)
{
maxPrice = p;
}
}
}
return maxPrice;
}
Inefficient, time complexity O (n- 2 )
2. traverse
traverse the array again, calculated for each day until the minimum stock price and maximum profit.
Diagram
int maxProfit(vector<int>& prices) {
if (prices.size() < 2)
{
return 0;
}
int minPrice = INT_MAX;
int maxPrice = 0;
for(int i = 0; i < prices.size(); i++)
{
if(minPrice > prices[i])
{
minPrice = prices[i];
}
else if(prices[i] - minPrice > maxPrice)
{
maxPrice = prices[i] - minPrice;
}
}
return maxPrice;
}
reference
