Best Time to Buy and Sell Stock II

Leetcode-Algorithm-Greedy-122

题目：

Say you have an array for which the ith 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).
（假定有一个数组，数组中的元素表示数组索引当天的股票价格。设计一个算法找到最大的利润。你可以完成尽可能多交易，可以当天买，当天卖，但是同一时间段不能发生多次交易。）

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题解：
方法：（贪心算法）
要想得到最大的收益，那么从第一天开始，在股价下降前就卖出股票。因此，用bday记录买入时间，sday记录卖出时间。起始时，bday=sday，然后用指针遍历数组。当下一天i的股价上升时，令sday=i；直到股价下降，就在sday卖出，在重置bday和sday于下降股价当天。因此，若股价下跌，就当天买卖，确保不会亏本。

class Solution {public:    int maxProfit(vector<int>& prices) {        if (prices.empty()) return 0;        int bday = 0;        int sday = 0;        int all = 0;        for (int i = 1; i < prices.size(); ++i) {            if (prices[i] >= prices[sday])                sday = i;            else {                all += prices[sday] - prices[bday];                bday = sday = i;            }        }        all += prices[sday] - prices[bday];        return all;    }};

分析：
因为只需遍历数组一次，所以时间复杂度为O(n)。