买卖股票的最佳时机4
昨天3的升级版,多一个奇偶判定。
            
            
              cpp
              
              
            
          
          class Solution {
public:
    int maxProfit(int k, vector<int>& prices) {
        if (prices.size() == 0) return 0;
        vector<vector<int>> dp(prices.size(), vector<int>(2 * k + 1, 0));
        for (int j = 1; j < 2 * k; j += 2) {
            dp[0][j] = -prices[0];
        }
        for (int i = 1;i < prices.size(); i++) {
            for (int j = 0; j < 2 * k - 1; j += 2) {
                dp[i][j + 1] = max(dp[i - 1][j + 1], dp[i - 1][j] - prices[i]);
                dp[i][j + 2] = max(dp[i - 1][j + 2], dp[i - 1][j + 1] + prices[i]);
            }
        }
        return dp[prices.size() - 1][2 * k];
    }
};
        买卖股票的最佳时机含冷冻期
            
            
              cpp
              
              
            
          
          class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        if (n == 0) return 0;
        vector<vector<int>> dp(n, vector<int>(4, 0));
        dp[0][0] -= prices[0]; // 持股票
        for (int i = 1; i < n; i++) {
            dp[i][0] = max(dp[i - 1][0], max(dp[i - 1][3] - prices[i], dp[i - 1][1] - prices[i]));
            dp[i][1] = max(dp[i - 1][1], dp[i - 1][3]);
            dp[i][2] = dp[i - 1][0] + prices[i];
            dp[i][3] = dp[i - 1][2];
        }
        return max(dp[n - 1][3], max(dp[n - 1][1], dp[n - 1][2]));
    }
};