LeetCode 122. Best Time to Buy and Sell Stock II [Easy] [C++] 解題筆記

Say you have an array prices for which the i^th element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete as many transactions as you like (i.e., buy one and sell one share of the stock multiple times).

Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).

Example 1:

Input: [7,1,5,3,6,4]
Output: 7
Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
             Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.

Example 2:

Input: [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
             Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
             engaging multiple transactions at the same time. You must sell before buying again.

Example 3:

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

想法:

    這題就是 greedy 的方式，從左至右遍歷，只要當前價格比前一個時間點還高就賣，然後將價差累加起來即可。

Complexity: O(n) time

完整程式碼:

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