LeetCode

 

 

Description

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 (ie, buy one and sell one share of the stock multiple times). However, you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

Solution

思路：只要列表中任意两个连续股价出现上涨趋势，就累积对应的增幅；对于股价下降的情况，则忽略之。

python

class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        length = len(prices)
        if (length == 0 or length == 1):
            return 0

        cur_diff = 0
        sum_profit = 0
        for i in range(1, length):
            cur_diff = prices[i] - prices[i-1]
            if cur_diff > 0:
                sum_profit = sum_profit + cur_diff

        return sum_profit

cpp

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int length = prices.size();
        if (length == 0 || length == 1)
            return 0;

        int sum_profit = 0;
        int cur_diff = 0;
        for (size_t i=1; i<length; ++i)
        {
            cur_diff = prices.at(i) - prices.at(i-1);
            if (cur_diff > 0)
            {
                sum_profit += cur_diff;
            }
        }

        return sum_profit;
    }
};