(贪心) leetcode 122. Best Time to Buy and Sell Stock II

Say you have an array for which the ith 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.

-----------------------------------------------------------------------------------------------------------------

源自leetcode 的官方题解。从第三个题解得到了启发。

这个题是贪心题，可以比较相邻的数的差，如果小于零，就是0，一次遍历就行。试想，从这个图中，我们要求的A和B的和，那么其他的下降的部分我们可以忽略，于是，这个忽略的部分用0代替，而上升的部分，则是累加。

C++代码：

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