121 Best Time to Buy and Sell Stock 買賣股票的最佳時機

Description:
Say you have an array for which the ith element is the price of a given stock on day i.

If you were only permitted to complete at most one transaction (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Note that you cannot sell a stock before you buy one.

Example:

Example 1:
Input: [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.
Not 7-1 = 6, as selling price needs to be larger than buying price.

Example 2:
Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

題目描述:
給定一個數組，它的第 i 個元素是一支給定股票第 i 天的價格。

如果你最多只允許完成一筆交易（即買入和賣出一支股票），設計一個算法來計算你所能獲取的最大利潤。

注意你不能在買入股票前賣出股票。

示例:

示例 1:
輸入: [7,1,5,3,6,4]
輸出: 5
解釋: 在第 2 天（股票價格 = 1）的時候買入，在第 5 天（股票價格 = 6）的時候賣出，最大利潤 = 6-1 = 5 。
注意利潤不能是 7-1 = 6, 因為賣出價格需要大于買入價格。

示例 2:
輸入: [7,6,4,3,1]
輸出: 0
解釋: 在這種情況下, 沒有交易完成, 所以最大利潤為 0。

思路:

動態規劃
前 i天的最大收益 = max[前 i天的價格 - min(前 i天價格)]
時間復雜度O(n), 空間復雜度O(1)

代碼:
C++:

class Solution 
{
public:
    int maxProfit(vector<int>& prices) 
    {
        int result = 0, purchase = 1e5;
        for (int i = 0; i < prices.size(); i++) 
        {
            if (prices[i] < purchase) purchase = prices[i];
            if (prices[i] - purchase > result) result = prices[i] - purchase;
        }
        return result;
    }
};

Java:

class Solution {
    public int maxProfit(int[] prices) {
        int result = 0, purchase = (1 << 31) - 1;
        for (int i = 0; i < prices.length; i++) {
            if (prices[i] < purchase) purchase = prices[i];
            if (prices[i] - purchase > result) result = prices[i] - purchase;
        }
        return result;
    }
}

Python:

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        result, purchase = 0, (1 << 31) - 1
        for price in prices:
            purchase, result = min(price, purchase), max(price - purchase, result)
        return result