代码随想录算法训练营第35天 | 第九章 动态规划08

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今日记录

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121. 买卖股票的最佳时机

Leetcode链接

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

122.买卖股票的最佳时机II

Leetcode链接

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

123.买卖股票的最佳时机III

Leetcode链接

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        vector<vector<int>> dp(prices.size(), vector<int>(5, 0));
        dp[0][1] = -prices[0];
        dp[0][3] = -prices[0];
        for (int i = 1; i < prices.size(); i++) {
            dp[i][0] = dp[i - 1][0];
            dp[i][1] = max(dp[i - 1][1], dp[i - 1][0] - prices[i]);
            dp[i][2] = max(dp[i - 1][2], dp[i - 1][1] + prices[i]);
            dp[i][3] = max(dp[i - 1][3], dp[i - 1][2] - prices[i]);
            dp[i][4] = max(dp[i - 1][4], dp[i - 1][3] + prices[i]);
        }
        return dp[prices.size() - 1][4];
    }
};

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总结

要思考清楚dp数组的含义！！