算法D48 | 动态规划10 | 121. 买卖股票的最佳时机 122.买卖股票的最佳时机II-优快云博客

本文链接：https://blog.youkuaiyun.com/memolaner/article/details/136790077

本文介绍了两个股票交易问题，分别是买卖股票的最佳时机和其升级版II，通过Python和C++代码展示了动态规划的解决方案，每个问题都附有B站视频讲解链接。

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股票问题是一个动态规划的系列问题，今日安排的题目不多，大家可以慢慢消化。

121. 买卖股票的最佳时机

视频讲解：https://www.bilibili.com/video/BV1Xe4y1u77q

https://programmercarl.com/0121.%E4%B9%B0%E5%8D%96%E8%82%A1%E7%A5%A8%E7%9A%84%E6%9C%80%E4%BD%B3%E6%97%B6%E6%9C%BA.html

Python:

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        n = len(prices)
        dp = [[0, 0] for _ in range(n)]
        dp[0][0] = -prices[0]
        dp[0][1] = 0
        for i in range(1, n):
            dp[i][0] = max(dp[i-1][0], -prices[i]) # 持有股票所得现金
            dp[i][1] = max(dp[i-1][1], prices[i]+dp[i-1][0]) # 不持有股票所得现金
        return dp[n-1][1]

C++:

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        vector<vector<int>> dp(n, vector<int>(2));
        dp[0][0] = -prices[0];
        dp[0][1] = 0;
        for (int i=1; i<n; i++) {
            dp[i][0] = max(dp[i-1][0], -prices[i]);
            dp[i][1] = max(dp[i-1][1], prices[i]+dp[i-1][0]);
        }
        return dp[n-1][1];
    }
};

122.买卖股票的最佳时机II

视频讲解：https://www.bilibili.com/video/BV1D24y1Q7Ls

https://programmercarl.com/0122.%E4%B9%B0%E5%8D%96%E8%82%A1%E7%A5%A8%E7%9A%84%E6%9C%80%E4%BD%B3%E6%97%B6%E6%9C%BAII%EF%BC%88%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%EF%BC%89.html

Python:

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

C++:

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        vector<vector<int>> dp(n, vector<int>(2, 0));
        dp[0][0] = -prices[0];
        dp[0][1] = 0;
        for (int i=1; i<n; i++) {
            dp[i][0] = max(dp[i-1][0], dp[i-1][1]-prices[i]);
            dp[i][1] = max(dp[i-1][1], dp[i-1][0]+prices[i]);
        }
        return dp[n-1][1];
    }
};