LeetCode 122 - Best Time to Buy and Sell Stock II

Difficulty: medium

Stock Series:

Part 1: LeetCode 121 - Best Time to Buy and Sell Stock
Part 2: LeetCode 122 - Best Time to Buy and Sell Stock II
Part 3: LeetCode 123 - Best Time to Buy and Sell Stock III
Part 4: LeetCode 188 - Best Time to Buy and Sell Stock IV
Part 5: LeetCode 309 - Best Time to Buy and Sell Stock with Cooldown
Part 6: LeetCode 714 - Best Time to Buy and Sell Stock with Transaction Fee

Problem Description

English (Best Time to Buy and Sell Stock II)

You are given an integer array prices where prices[i] is the price of a given stock on the $i^{th}$ day.

On each day, you may decide to buy and/or sell the stock. You can only hold at most one share of the stock at any time. However, you can buy it then immediately sell it on the same day.

Find and return the maximum profit you can achieve.

Example 1:

Input: prices = [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.
Total profit is 4 + 3 = 7.

Example 2:

Input: prices = [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.
Total profit is 4.

Example 3:

Input: prices = [7,6,4,3,1]
Output: 0
Explanation: There is no way to make a positive profit, so we never buy the stock to achieve the maximum profit of 0.

Constraints:

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

Chinese (买卖股票的最佳时机 II)

给你一个整数数组 prices ，其中 prices[i] 表示某支股票第 i 天的价格。

在每一天，你可以决定是否购买和/或出售股票。你在任何时候 最多 只能持有 一股 股票。你也可以先购买，然后在 同一天 出售。

返回 你能获得的 最大 利润 。

示例 1：

输入：prices = [7,1,5,3,6,4]
输出：7
解释：在第 2 天（股票价格 = 1）的时候买入，在第 3 天（股票价格 = 5）的时候卖出, 这笔交易所能获得利润 = 5 - 1 = 4 。
     随后，在第 4 天（股票价格 = 3）的时候买入，在第 5 天（股票价格 = 6）的时候卖出, 这笔交易所能获得利润 = 6 - 3 = 3 。
     总利润为 4 + 3 = 7 。

示例 2：

输入：prices = [1,2,3,4,5]
输出：4
解释：在第 1 天（股票价格 = 1）的时候买入，在第 5 天 （股票价格 = 5）的时候卖出, 这笔交易所能获得利润 = 5 - 1 = 4 。
     总利润为 4 。

示例 3：

输入：prices = [7,6,4,3,1]
输出：0
解释：在这种情况下, 交易无法获得正利润，所以不参与交易可以获得最大利润，最大利润为 0 。

提示：

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

Solutions

Solution 1

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        if (n == 0) {
            return 0;
        }

        int res = 0;
        int prev = prices[0];
        for (int i = 1; i < n; ++i) {
            int curr = prices[i];
            if (curr > prev) {
                res += curr - prev;
            }

            prev = curr;
        }

        return res;
    }
};

Solution 2

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int res = 0;
        for (int i = 1, n = prices.size(); i < n; ++i) {
            if (int diff = prices[i] - prices[i - 1]; diff > 0) {
                res += diff;
            }
        }

        return res;
    }
};

Solution 3

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        if (n == 0) {
            return 0;
        }

        vector<int> has(n), not_has(n);
        has[0] = -prices[0];

        for (int i = 1; i < n; ++i) {
            has[i] = max(has[i - 1], not_has[i - 1] - prices[i]);
            not_has[i] = max(not_has[i - 1], has[i - 1] + prices[i]);
        }

        return max(has[n - 1], not_has[n - 1]);
    }
};