本文主要是介绍[leetcode] 122. Best Time to Buy and Sell Stock II，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
这是一个非常简单贪心问题，贪心问题的解决方法一般都非常简单，但是要想到这个方法真的挺不容易的：
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 (i.e., buy one and sell one share of the stock multiple times).
Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).
Example 1:
Input: [7,1,5,3,6,4] Output: 7 Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.Example 2:
Input: [1,2,3,4,5] Output: 4 Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.Note that you cannot buy on day 1, buy on day 2 and sell them later, as you areengaging multiple transactions at the same time. You must sell before buying again.Example 3:
Input: [7,6,4,3,1] Output: 0 Explanation: In this case, no transaction is done, i.e. max profit = 0.
代码为：
class Solution {
public:int maxProfit(vector<int>& prices) {int maxP = 0;if(prices.size() == 0)return maxP;for(int i = 0; i < prices.size() - 1; i++){if(prices[i] < prices[i + 1])maxP += prices[i + 1] - prices[i];}return maxP;}
};看起来是不是很简单，用到了贪心算法的核心思想，用局部最优的解法解决整体问题，只要股票升了就卖，按照这个解法得到的结果是要求得的最大利润，但是这个过程却不是符合本题买卖股票的中间过程的，中间过程类似于当股票在一个极小区间内属于最小时就买入，在一个极小区间内属于最大时就抛售，在这两段过程中要保证获得利润最大，（实际上这样理解还是有纰漏，这也是贪心算法的困难所在，想到贪心算法的解决办法是非常困难的，需要大量的题目训练才可以），这样分解的过程就是获得最大利润的过程。
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