122. Best Time to Buy and Sell Stock II 股票最佳买入和卖出时间

允许股票买入和卖出多次，并且要求只能在卖出后，才能买入。

这道题和 121. Best Time to Buy and Sell Stock 股票最佳买入和卖出时间 不同，121这道题相对比较简单，只要求，买卖一次。而这道题可以买卖多次，得到最大的收益。

例如：
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.

难度：【easy】 (虽然写着的难度系数不高，但是思考起来不是那么容易想到的)

方法一：Peak Valley Approach 谷峰法

每次取得一个 极低点valley 和 极高点peak。

    int maxProfit(vector& prices) {
        if (prices.size() <= 1) { return 0; } 
        
        int i = 0;
        int valley = prices[0];
        int peak = prices[0];
        int maxprofit = 0;
        
        while (i < prices.size() - 1) {
            while (i < prices.size() - 1 && prices[i] >= prices[i + 1])
                i++;
            valley = prices[i];
            
            while (i < prices.size() - 1 && prices[i] <= prices[i + 1])
                i++;
            peak = prices[i];
            
            maxprofit += peak - valley;
        }
        
        return maxprofit;
    }

方法二：是对方法一的简化版

    int maxProfit(vector& prices) {
        int maxprofit = 0;
        for (int i = 1; i < prices.size(); ++i) {
            if (prices[i] > prices[i - 1]) {
                maxprofit += prices[i] - prices[i - 1];
            }
        }
        
        return maxprofit;
    }