122. 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).

分析：在上一题的基础上，可以买卖多次股票，但是不能连续买股票，也就是说手上最多只能有一只股票（注意：可以在同一天卖出手上的股票然后再买进）

算法1：同上一题一样构建股票差价数组，把数组中所有差价为正的值加起来就是最大利润了。把解题思想更通俗地来说，就是不放过一丝一毫可以挣钱的机会。

class Solution {
public:
int maxProfit(vector<int> &prices) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
int len = prices.size();
if(len <= 1)return 0;
int res = 0;
for(int i = 0; i < len-1; i++)
if(prices[i+1]-prices[i] > 0)
res += prices[i+1] - prices[i];
return res;
}
};

9.15更新

II并没有限制总的买卖次数，只限制了不能同时买和卖。所以可以采用greedy的方法，来获得所有可能的正利润。以如下序列说明：

2 1 3 4 5 4 2 8 7

只要prices[i] - prices[i-1]>0，我们就在第i-1天买入，第i天抛出。这样可以包括所有可能赚取利润的区间。

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