力扣-数组-121 买卖股票的最佳时机

代码

超内存了

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

二维变一维，不超内存，超时间了

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int dp[prices.size()]; // i  ——  size()-1 最大的那一个 
        for(int i = 0;i < prices.size();i++){
            int last = 0;
            for(int j = i; j<prices.size();j++){
                if(i == j){
                    last = 0;
                }else{
                    int temp = max( max( max(0, prices[j]-prices[i]), last), prices[j]-prices[i+1]);
                    last = temp;
                }
            }
            dp[i] = last;
        }
        int res = 0;
        for(int i = 0; i< prices.size();i++){
            res = max(res, dp[i]);
        }
        return res;
    }
};

看了大佬的题解，才明白跟dp没啥关系，主要应用贪心了

最后附上基于此思想的代码

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