122 - Best Time To Buy And Sell Stock II

#medium

You are given an integer array prices where prices[i] is the price of a given stock on the ith day.

On each day, you may decide to buy and/or sell the stock. You can only hold at most one share of the stock at any time. However, you can buy it then immediately sell it on the same day.

Find and return the maximum profit you can achieve.

Example 1:
Input: prices = [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.
Total profit is 4 + 3 = 7.

Example 2:
Input: prices = [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.
Total profit is 4.

Example 3:
Input: prices = [7,6,4,3,1]
Output: 0
Explanation: There is no way to make a positive profit, so we never buy the stock to achieve the maximum profit of 0.

貪心算法

這道題目可能我們只會想，選一個低的買入，再選個高的賣，再選一個低的買入.....循環反覆。

如果想到其實最終利潤是可以分解的，那麼本題就很容易了！ (同天賣掉可以再買)

如何分解呢？

假如第0天買入，第3天賣出，那麼利潤為：prices[3] - prices[0]。

相當於(prices[3] - prices[2]) + (prices[2] - prices[1]) + (prices[1] - prices[0])。

此時就是把利潤分解為每天為單位的維度，而不是從0天到第3天整體去考慮！

那麼根據prices可以得到每天的利潤序列：(prices[i] - prices[i - 1]).....(prices[1] - prices[0])。

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

        return result;
    }
};