122. Best Time to Buy and Sell Stock II

Say you have an array 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.

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还是股票买卖题。这一题不限制交易次数，但还是要求必须先卖出再买入，同一天可以先卖出再买入。 贪心策略，只要当前的股价比前一天高，则在前一天买入，当天卖出，能赚一点是一点，然后把每天赚的累加起来。 可能稍微会担心是否能得到最优解，比如价格是1,2,900,901，如果只交易一次，最好在第一天买入，最后一天卖出，收益901-1=900，但是贪心策略也一样：(2-1)+(900-2)+(901-900)=(901-1)=900。所以贪心能得到最优解。完整代码如下：

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

本代码提交AC，用时9MS。