In this **Leetcode Best Time to Buy and Sell Stock II problem solution**, we have given an array of prices where prices[i] is the price of a given stock on an ith day. Find the maximum profit you can achieve. You may complete as many transactions as you like (i.e., buy one and sell one share of the stock multiple times).

## Problem solution in Python.

class Solution: def maxProfit(self, prices: List[int]) -> int: maxprofit=0 for i in range(1,len(prices)): if prices[i]>prices[i-1]: maxprofit+=prices[i]-prices[i-1] return maxprofit

## Problem solution in Java.

public int maxProfit(int[] prices) { int max_profit = 0, i = 0, current_profit; while(i < prices.length) { current_profit = 0; while (i+1 < prices.length && prices[i] > prices[i+1]) i++; if(i >= prices.length) return max_profit; current_profit = current_profit - prices[i]; while (i+1 < prices.length && prices[i] < prices[i+1]) i++; if(i >= prices.length) return max_profit; current_profit = current_profit + prices[i]; max_profit = max_profit + current_profit; i++; } return max_profit; }

## Problem solution in C++.

int maxProfit(vector<int> &prices) { int ret = 0; for (size_t p = 1; p < prices.size(); ++p) ret += max(prices[p] - prices[p - 1], 0); return ret; }

## Problem solution in C.

int maxProfit(int* a, int n){ int i=1, profit = 0; for(; i<n; i++){ if( a[i] - a[i-1] > 0){ profit += a[i] - a[i-1]; } } return profit; }

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