【Leetcode122】Best Time to Buy and Sell Stock II

最直观的想，首先这个问题的实质是求解所有的峰值与左侧的谷底之间的差值之和，但是一开始要判断初值，弄了个标志位。还有为了判断峰值还是谷底所产生的越界问题。无论是跟前面比还是跟后面比。以下是错误的思路：

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int siz = prices.size();
        bool on = 0;
        int sum = 0;
        int buy;
        for(int i = 0; i < siz; ++i){
            if(prices[i] < prices[i+1] && !on){
                buy = prices[i];
                on = true;
            }
            else if(prices[i] > prices[i+1] && on){
                sum += prices[i] - buy;
                on = false;
            }
        }
        return sum;
    }
};

陷入了困境

换个思路，不考虑可以买卖的次数，任意两天之间，只要盈利就进行交易，不管是不是峰值还是谷底！

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

但是时间复杂度比较高，以后再看