【代码随想录】day50

提示：文章写完后，目录可以自动生成，如何生成可参考右边的帮助文档

一、123买卖股票的最佳时机III

超时了。。。

class Solution {
public:
    int helper(vector<int>& prices, int start, int end) {
        int profit = 0;
        int buyPrice = prices[start];
        for (int i = start + 1; i < end; i ++) {
            profit = max(profit, prices[i] - buyPrice);
            buyPrice = min(buyPrice, prices[i]);
        }
        return profit;
    }

    int maxProfit(vector<int>& prices) {
        int result = 0;
        for (int i = 1; i < prices.size(); i ++) {
            if (prices[i] <= prices[i-1]) {
                continue;
            }
            result = max(result, helper(prices, 0, i) + helper(prices, i - 1, prices.size()));
        }
        return result;
    }
};

二维dp：

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][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]);
        }
        // for (auto vec: dp) {
        //     for (int num: vec) cout << num << " ";
        //     cout << endl;
        // }
        return dp[prices.size()-1][4];
    }
};

二、188买卖股票的最佳时机IV

上一道题举一反三即可，但上一道题的二维dp真滴想不到呀。。

class Solution {
public:
    int maxProfit(int k, vector<int>& prices) {
        vector<vector<int>> dp(prices.size(), vector<int>(1 + 2 * k, 0));
        for (int i = 1; i < dp[0].size(); i += 2) {
            dp[0][i] = -prices[0];
        }
        for (int i = 1; i < prices.size(); i ++) {
            for (int j = 1; j < dp[0].size(); j ++) {
                if (j % 2 == 1) {
                    dp[i][j] = max(dp[i-1][j], dp[i-1][j-1] - prices[i]);
                }
                else {
                    dp[i][j] = max(dp[i-1][j], dp[i-1][j-1] + prices[i]);
                }
            }
        }

        // for (auto vec: dp) {
        //     for (int num: vec) cout << num << " ";
        //     cout << endl;
        // }
        return dp[prices.size() - 1][2 * k];
    }
};