LeetCode [122] Best Time to Buy and Sell Stock II

122. Best Time to Buy and Sell Stock II

Easy

Say you have an array prices for which the ith element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete as many transactions as you like (i.e., buy one and sell one share of the stock multiple times).

Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).

Example 1:

Input: [7,1,5,3,6,4]
Output: 7
Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
             Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.

Example 2:

Input: [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
             Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
             engaging multiple transactions at the same time. You must sell before buying again.

Example 3:

Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

 

Constraints:

• 1 <= prices.length <= 3 * 10 ^ 4
• 0 <= prices[i] <= 10 ^ 4

 

Note

• buy_price: the potential buying price, initialized by prices[0]
• if 
• the current price prices[i]<=buy_price, update buy_price by prices[i]
• else (when prices[i]>buy_price)
• sell the stock when the price is about to decrease (or it's the last price)

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//C++: 8ms class Solution { public: int maxProfit(vector<int>& prices) { int maxP = 0, n = prices.size(); for(int i=1; i<n; ++i){ if(prices[i]>prices[i-1]) maxP += prices[i]-prices[i-1]; } return maxP; } }; //C++: 8ms class Solution { public: int maxProfit(vector<int>& prices) { int i = 0, n = prices.size(), profit = 0; while(i<n){ while(i<n-1 && prices[i]>=prices[i+1]) i++; if(i==n) return profit; int buy = prices[i]; while(i<n-1 && prices[i]<=prices[i+1]) i++; profit += prices[i]-buy; i++; } return profit; } }; //C++: 8ms class Solution { public: int maxProfit(vector<int>& prices) { int n = prices.size(); if(n<=1) return 0; int buy_price = prices[0]; int max_profit = 0; for(int i=1; i<n; ++i){ if(prices[i]<=buy_price) buy_price = prices[i]; else if(i==n-1 || prices[i]>=prices[i+1]){ max_profit += prices[i]-buy_price; if(i<n-1) buy_price = prices[i+1]; } } return max_profit; } }; //Java class Solution { public int maxProfit(int[] prices) { int r = 0; for(int i=1; i<prices.length; ++i) { if(prices[i]>prices[i-1]) r += prices[i]-prices[i-1]; } return r; } }