[LeetCode] 121. Best Time to Buy and Sell Stock

[frdmu]

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

You want to maximize your profit by choosing a single day to buy one stock and choosing a different day in the future to sell that stock.

Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return 0.

 

Example 1:

Input: prices = [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.
Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.

Example 2:

Input: prices = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transactions are done and the max profit = 0.

Constraints:

• 1 <= prices.length <= 105
• 0 <= prices[i] <= 104

 

解法一：
贪心法。遍历数组，计算当天能达到的最大利润，取其中的最大值。代码如下：

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int mi = INT_MAX, res = 0;
        
        for (auto price: prices) {
            if (price < mi) {
                mi = price;
            }
            res = max(res, price - mi);
        }

        return res;
    }
};

 

解法二：
动态规划。

dp[i][0]表示第i天持有股票所获得的的最大现金, dp[i][1]表示第i天不持有股票所获得的的最大现金。

对于dp[i][0]，它有两种来源：

• 前一天已经持有股票，则dp[i][0] = dp[i-1][0]
• 前一天未持有股票，则第i天才买入，则dp[i][0] = -prices[i]
  综上：dp[i][0] = max(dp[i-1][0], -prices[i])

对于dp[i][1]，它有两种来源：

• 前一天也未持有股票，则dp[i][1] = dp[i-1][1]
• 前一天持有股票，则第i天才卖出，则dp[i][1] = prices[i] + dp[i-1][0]
  综上：dp[i][1] = max(dp[i-1][1], prices[i]+dp[i-1][0])

图示如下：

代码如下：

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];

        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], dp[i-1][0]+prices[i]);
        }

        return dp[n-1][1];
    }
};

 

解法三：
动态规划。利用统一的框架:动态规划总结: 股票买卖问题进行推导，代码如下：

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        vector<vector<int>> dp(n, vector<int>(2, 0));

        dp[0][1] = -prices[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], - prices[i]);
        }

        return dp[n-1][0];
    }
};

Refer:
团灭 LeetCode 股票买卖问题