LeetCode 121. Best Time to Buy and Sell Stock 解題紀錄
Contents
題目
Problem
You are given an array prices where prices[i] is the price of a given stock on the $i^{th}$ day.
You want to maximize your profit by choosing a single day to buy one stock and choosing a different day in the future to sell that stock.
Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return 0.
Example 1:
Input: prices = [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5. Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.
Example 2:
Input: prices = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transactions are done and the max profit = 0.
Constraints:
- 1 $\leq$ prices.length $\leq 10^5$
- 0 $\leq$ prices[i] $\leq 10^4$
想法
Info
解法
解法一
int maxProfit(vector<int>& prices) {
if(prices.size()<2) return 0;
auto buy = INT_MAX, profit = 0;
for(auto i:prices)
{
buy = min(i, buy);
profit = max(profit, i-buy);
}
return profit;
}
解法二
int maxProfit(vector<int>& prices) {
int profit = 0, buy = prices[0];
if(prices.size() < 2) return 0;
for(auto i=1; i<prices.size(); ++i)
{
if(prices[i] > prices[i-1])
{
profit = max(profit, prices[i] - buy);
}
else
{
buy = min(buy, prices[i]);
}
}
return profit;
}
- Time complexity: $\mathcal{O}(n)$.
- Space complexity: $\mathcal{O}(n)$.