Leetcode#122. Best Time to Buy and Sell Stock II（股票多次买卖的最佳时间--贪心）

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).

题意

http://blog.csdn.net/xunalove/article/details/79173830与上一题题意类似，121要求是一次买卖，这道题可以进行多次买卖。

题解

先拿一组数据分析：

1 2 7 4 5 6

最大值：（7-1） + （6-4 ) = 6 + 2 = 8

也可以拆分成：( 2-1) + (7-2) + (5-4) + (6-5) = 1 + 5 + 1 + 1 = 8

即全局最优解 = 局部最优解之和（贪心思想）

Python语言

class Solution(object):
def maxProfit(self, prices):
"""
:type prices: List[int]
:rtype: int
"""
profit = 0
for index in range(1, len(prices)):
if prices[index]-prices[index-1] > 0:
profit = profit + (prices[index] - prices[index-1])
return profit

C++语言

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