算法设计Week5 LeetCode Algorithms Problem #122 Best Time to Buy and Sell Stock II

题目描述：

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).

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解法一：

考虑模拟整个股票的买入卖出过程：如果没有持股且第二天股票价格比今天高，则买入；如果持股且第二天股票价格比今天低，则卖出；其他情况不进行考虑。到最后一天时，如果持股且股票价格比买入价格高，则卖出。算法复杂度O(n)，空间复杂度O(1)。实现的代码如下：

class Solution {public:    int maxProfit(vector<int>& prices) {        if(prices.empty()) return 0;        int profit = 0;        int buy_price;        bool hold = false;        for(int i = 0; i < prices.size() - 1; i++){            if(prices[i] < prices[i + 1] && !hold){ // buy                hold = true;                buy_price = prices[i];            }else if(prices[i] > prices[i + 1] && hold){  // sell                hold = false;                profit += prices[i] - buy_price;            }        }        // last day        if(hold && prices.back() > buy_price){  // sell            hold = false;            profit += prices.back() - buy_price;        }        return profit;    }};

解法二：

使用贪心算法，不考虑何时进行交易，整个股票中全部的利润都要获得才能达到最大利润。这样，只要将数组中前后两项和为正（即能够产生利润的部分）相加，就能够得解。算法复杂度O(n)，空间复杂度O(1)。实现的代码如下：

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