The maximum profit (leetcode 122) p64 conduct an unlimited number of stock trading

A: problem-solving ideas

Method one: constantly looking for the minimum and maximum stock, and the smallest is worth the time to buy and sell at the maximum worth the time. Time: O (n), Space: O (1)

Method Two: greedy, just a day after the value is greater than the value of the previous day, then be traded. Time: O (n), Space: O (n)

Two: Complete code examples (C ++ version and the Java version)

Method one C ++:

class Solution {
public:
    int maxProfit(vector<int>& prices) 
    {
        if (prices.size() == 0) return 0;
        int n = prices.size();
        int i = 0, buy = 0, sell = 0,maxProfit=0;

        while (i < n - 1)
        {
            while (i < n - 1 && prices[i + 1] <= prices[i]) i++;
            buy = prices[i];
            while (i < n - 1 && prices[i + 1] >= prices[i]) i++;
            sell = prices[i];

            maxProfit += (sell-buy);
        }

        return maxProfit;
    }
};

Method One Java:

class Solution {
    public int maxProfit(int[] prices)
    {
           if(prices==null||prices.length==0) return 0;
           int maxProfit=0,i=0,buy=0,sell=0;
           int n=prices.length;
           
           while(i<n-1)
           {
               while(i<n-1 && prices[i+1]<=prices[i]) i++;
               buy=prices[i];
               while(i<n-1 && prices[i+1]>=prices[i]) i++;
               sell=prices[i];
               
               maxProfit+=(sell-buy);
           }
           
           return maxProfit;
    }
}

Method Two C ++:

class Solution {
public:
    int maxProfit(vector<int>& prices) 
    {
        if (prices.size() == 0) return 0;
        int maxProfit=0;

        for (int i = 1; i < prices.size(); i++)
        {
            if (prices[i] > prices[i - 1])
                maxProfit += (prices[i]-prices[i-1]);
        }

        return maxProfit;
    }
};

Method Two Java:

class Solution {
    public int maxProfit(int[] prices)
    {
           if(prices==null||prices.length==0) return 0;
           int maxProfit=0;
           
           for(int i=1;i<prices.length;i++)
           {
               if(prices[i]>prices[i-1])
                   maxProfit+=(prices[i]-prices[i-1]);
           }

           return maxProfit;
    }
}