Leetcode Best Time to Buy and Sell Stock with Cooldown

题意:给出每个时刻股票的价格,每天只能买入或卖出,买入或卖出后第二天不能交易。

思路:DP,dp[i][j] 表示在第i天买入,第j天卖出的最大利润,转态方程为:dp[i][j] = max(dp[i][j - 1],  max(dp[i - 2][k])。

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        vector<int> dp(prices.size(), 0);
        int max_pro = 0;
        vector<int> dp_pre(prices.size(), 0);
        
        for(int i = 0; i < prices.size(); ++ i) {
            for(int j = i; j < prices.size(); j ++) {
                int temp_max = 0;
                if(i - 2 >= 0) temp_max = dp_pre[i - 2];
               
                if(prices[j] - prices[i] > 0) dp[j] = prices[j] - prices[i] + temp_max;
                else dp[j] = temp_max;
                if(j - 1 > 0) dp[j] = max(dp[j], dp[j - 1]);
                max_pro = max(max_pro, dp[j]);
                dp_pre[j] = max(dp_pre[j], dp[j]);
                //printf("%d ", dp[i][j]);
            }
            //cout << endl;
        }
        
        return max_pro;
    }
    
};1


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转载自blog.csdn.net/markpen/article/details/59483509