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LeetCode 122. Best Time to Buy and Sell Stock II #35

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Woodyiiiiiii opened this issue Apr 27, 2020 · 0 comments
Open

LeetCode 122. Best Time to Buy and Sell Stock II #35

Woodyiiiiiii opened this issue Apr 27, 2020 · 0 comments

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@Woodyiiiiiii
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Say you have an array prices 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.

Constraints:

  • 1 <= prices.length <= 3 * 10 ^ 4
  • 0 <= prices[i] <= 10 ^ 4

如果今日比明日价格低,那么就今日买入,明日卖出。遍历数组。

class Solution {
    public int maxProfit(int[] prices) {
        int res = 0;
        for (int i = 0; i < prices.length - 1; ++i) {
            if (prices[i] < prices[i + 1])
                res += prices[i + 1] - prices[i];
        }
        
        return res;
    }
}

参考资料:

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