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：
從前向后遍歷數(shù)組，只要當(dāng)天的價(jià)格高于前一天的價(jià)格，就算入收益。時(shí)間復(fù)雜度O(n)岛抄，空間復(fù)雜度O(1).
具體代碼如下：
class Solution {
public:
int maxProfit(vector<int>& prices) {
int res = 0, n = prices.size();
for (int i = 0; i < n - 1; ++i) {
if (prices[i] < prices[i + 1]) {
res += prices[i + 1] - prices[i];
}
}
return res;
}
};
