README_EN.md

122. Best Time to Buy and Sell Stock II

中文文档

Description

You are given an array prices where prices[i] is the price of a given stock on the ith day.

Find the maximum profit you can achieve. 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 simultaneously (i.e., you must sell the stock before you buy again).

 

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.

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.
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: prices = [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 * 104
• 0 <= prices[i] <= 104

Solutions

Greedy or Dynamic Programming.

Python3

Greedy:

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        res = 0
        for i in range(1, len(prices)):
            t = prices[i] - prices[i - 1]
            res += max(t, 0)
        return res

Dynamic Programming:

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        f1, f2 = -prices[0], 0
        for price in prices[1:]:
            f1 = max(f1, f2 - price)
            f2 = max(f2, f1 + price)
        return f2

Java

Greedy:

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

Dynamic Programming:

class Solution {
    public int maxProfit(int[] prices) {
        int f1 = -prices[0], f2 = 0;
        for (int i = 1; i < prices.length; ++i) {
            f1 = Math.max(f1, f2 - prices[i]);
            f2 = Math.max(f2, f1 + prices[i]);
        }
        return f2;
    }
}

TypeScript

function maxProfit(prices: number[]): number {
    let ans = 0;
    for (let i = 1; i < prices.length; i++) {
        ans += Math.max(0, prices[i] - prices[i - 1]);
    }
    return ans;
}

C++

Greedy:

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

Dynamic Programming:

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int f1 = -prices[0], f2 = 0;
        for (int i = 1; i < prices.size(); ++i) {
            f1 = max(f1, f2 - prices[i]);
            f2 = max(f2, f1 + prices[i]);
        }
        return f2;
    }
};

Go

Greedy:

func maxProfit(prices []int) int {
	res := 0
	for i := 1; i < len(prices); i++ {
		t := prices[i] - prices[i-1]
		if t > 0 {
			res += t
		}
	}
	return res
}

Dynamic Programming:

func maxProfit(prices []int) int {
	f1, f2 := -prices[0], 0
	for _, price := range prices[1:] {
		f1 = max(f1, f2-price)
		f2 = max(f2, f1+price)
	}
	return f2
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

C#

Greedy:

public class Solution {
    public int MaxProfit(int[] prices) {
        int res = 0;
        for (int i = 1; i < prices.Length; ++i)
        {
            int t = prices[i] - prices[i - 1];
            res += Math.Max(t, 0);
        }
        return res;
    }
}

Dynamic Programming:

public class Solution {
    public int MaxProfit(int[] prices) {
        int f1 = -prices[0], f2 = 0;
        for (int i = 1; i < prices.Length; ++i)
        {
            f1 = Math.Max(f1, f2 - prices[i]);
            f2 = Math.Max(f2, f1 + prices[i]);
        }
        return f2;
    }
}

...