算法设计Week9 LeetCode Algorithms Problem #121. Best Time to Buy and Sell Stock

题目描述：

Say you have an array for which the ith element is the price of a given stock on day i.

If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.

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题目解析：

题目要求其实是找出数组中位置较后和位置较前的两个数的最大差值。如果使用maxprofit[i]表示到第i天为止，进行一次买卖可以获取的最大利润，prices[i]表示第i天的股票价格，priceIn表示买入的股票价格，则有：

maxprofit[i]=max(maxprofit[i−1],prices[i]−priceIn).

而如果prices[i]<priceIn，则需要更新priceIn。具体代买如下所示，代码的时间复杂度为O(n)，空间复杂度为O(1)。

class Solution {public:    int maxProfit(vector<int>& prices) {        if(prices.empty()) return 0;        int profit = 0;        int price_in = prices[0];        for(int i = 1; i < prices.size(); i++){              profit = max(profit, prices[i] - price_in);            if(prices[i] < price_in){                price_in = prices[i];            }        }        return profit;    }};