122. Best Time to Buy and Sell Stock II - easy
문제
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).
제한사항
입출력 예
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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.
풀이
- Array, Greedy
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class Solution {
public:
int maxProfit(vector<int>& prices) {
int answer = 0;
// 가격이 증가하고있을때,
// 현재 가격이 전날의 가격보다 큰 현재 가격을 기준으로
// 현재 가격 - 전날 가격
// 을 계속해서 더해줌
for(auto i = 1 ; i < prices.size() ; ++i){
if(prices[i-1] < prices[i])
answer += prices[i] - prices[i-1];
}
return answer;
}
};
풀이참고
