題目鏈接
tag:
- Easy啊送；
question：
??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 are
engaging 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.
思路：
??本題跟之前那道Best Time to Buy and Sell Stock 最佳的買賣股票時(shí)間很類似房资，但都比較容易挨厚。這道題由于可以無限次買入和賣出铝耻。我們都知道炒股想掙錢當(dāng)然是低價(jià)買入高價(jià)拋出慈格，那么這里我們只需要從第二天開始仍秤，如果當(dāng)前價(jià)格比之前價(jià)格高失暂，則把差值加入利潤中彼宠，因?yàn)槲覀兛梢宰蛱熨I入，今日賣出弟塞，若明日價(jià)更高的話凭峡，還可以今日買入，明日再拋出决记。以此類推摧冀，遍歷完整個(gè)數(shù)組后即可求得最大利潤。代碼如下：
class Solution {
public:
int maxProfit(vector<int>& prices) {
if (prices.empty()) return 0;
int res = 0;
for (int i=0; i<prices.size()-1; ++i) {
if (prices[i] < prices[i+1]) {
res += prices[i+1] - prices[i];
}
}
return res;
}
};
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