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122. Best Time to Buy and Sell Stock II

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).

题解:

这道题的意思就是说给你一串数字,代表着一段时间内的交易价格,你可以选择在这一天做买入操作或者卖出操作,不限制操作次数,但是卖出之前必须有买入,问获得的最大收益是多少。这道题简单在不限操作。看上去好像是一个很复杂的排列组合,但是其实转换成贪心问题,我们可以这样思考:即每一天的最大收益是多少。再细致一点,无非是,今天要么收益要么没有收益。如果收益了,那就是今天的最大收益,实际上是比昨天的最大收益多了一个今天和昨天的价格差。

代码:

 1 class Solution {
 2 public:
 3     int maxProfit(vector<int>& prices) {
 4         int pro = 0;
 5         for(int i = 1; i < prices.size();i++)
 6         {
 7             pro = pro>pro+prices[i]-prices[i-1]?pro:pro+prices[i]-prices[i-1];
 8         }
 9         return pro;
10     }
11 };

 

122. Best Time to Buy and Sell Stock II