[leetcode] Best Time to Buy and Sell Stock II

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).

给定一个数组，第i个元素代表第i天的股票价格，现在可以进行无数次交易，但是在同一天中不能交易多次，求最大利润。

依次计算前后两天的差值作为利润，如果利润大于零，就加入到最大收益中，否则跳过。时间复杂度o(n)。

代码如下：

 1 class Solution {
 2 public:
 3     int maxProfit(vector<int> &prices) {
 4         // Start typing your C/C++ solution below
 5         // DO NOT write int main() function
 6         int result=0;
 7         if(prices.size()==0||prices.size()==1) return 0;
 8         for(int i=0;i<prices.size()-1;i++)
 9         {
10             int profit=prices[i+1]-prices[i];
11             if(profit>0)
12             {
13                 result+=profit;
14             }
15         }
16         return result;
17     }
18 };

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[leetcode] Best Time to Buy and Sell Stock II

原文：http://www.cnblogs.com/jostree/p/3704367.html