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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).
解题思路：
贪心算法：只要当前值比前一个值小，就更新利润。
1 class Solution { 2 public: 3 int maxProfit(vector<int>& prices) { 4 if (prices.size() == 0 || prices.size() == 1) { 5 return 0; 6 } 7 8 int min_index = 0; 9 int profit = 0; 10 11 for (int i = 1; i < prices.size(); ++i) { 12 if (prices[i] < prices[i - 1]) { 13 profit += prices[i - 1] - prices[min_index] > 0? prices[i - 1] - prices[min_index] : 0; 14 min_index = i; 15 } 16 } 17 if (prices[min_index] < prices[prices.size() - 1]) { 18 profit += prices[prices.size() - 1] - prices[min_index]; 19 } 20 21 return profit; 22 } 23 };
[LeetCode]Best Time to Buy and Sell Stock II
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原文地址：http://www.cnblogs.com/skycore/p/5361256.html
