题解 | #买卖股票的最好时机(一)#

• 计算收益：prices[i]-prices[i-1]；
• 计算连续收益的最大值。

class Solution {
public:
    /**
     * 
     * @param prices int整型vector 
     * @return int整型
     */
    int maxProfit(vector<int>& prices) {
        // write code here
        int result = 0;
        int count = 0;
        for (int i = 1; i < prices.size(); i++) {
            count += prices[i] - prices[i - 1];
            if (count > result) result = count;
            if (count < 0) count = 0;
        }
        return result;
    }
};