代码随想录算法训练营第五十天| LeetCode123.买卖股票的最佳时机III 188.买卖股票的最佳时机IV

123.买卖股票的最佳时机III

题目：力扣

class Solution {
public:
    int maxProfit(vector& prices) {
        vector > dp(prices.size(),vector(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];
    }
};

188.买卖股票的最佳时机IV

题目：力扣

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

总结

题型：买卖股票