"Leetcode" best time to Buy and Sell Stock II

Source: Internet
Author: User

Test instructions

Say you has an array for which the i-th element is the price of a given-stock on day I.

Design an algorithm to find the maximum profit. You could complete as many transactions as (ie, buy one and sell one share of the stock multiple times). However, engage in multiple transactions for the same time (ie, you must sell the stock before you buy again).


The maximum benefit that can be obtained by asking to buy or sell a stock multiple times. So, in the sequence of all the stock price movements, I just need to find out all the ascending sequence, buy the stock at the beginning of each monotonous ascending sequence, sell the stock at the end, and get the profit of the stock price rise.

So, how are these ascending sequences calculated? In fact, it is very simple, because each ascending sequence is composed of a two-day stock price increase, that is, if there is a 1 3 6 7 rise sequence, you as long as (1,3), (3,6), (6,7) small sequence composition, and then profit is the difference between each cell. Simply put, just meet the current price is higher than the previous day, add to the total profit, the final value is the answer.



Class Solution {public:    int maxprofit (vector<int> &prices) {        int ans = 0;        int n = prices.size ();        for (int i = 1;i < N;++i)        {            if (Prices[i] > prices[i-1])            {                ans + = prices[i]-prices[i-1];            }        }        return ans;}    ;

Class solution:    # @param prices, a list of integer    # @return An integer    def maxprofit (self, prices):        n = l En (prices)        ans = 0 for        i in range (1,n):            if prices[i] > prices[i-1]:                ans + = prices[i]-prices[i-1]                        return ans

"Leetcode" best time to Buy and Sell Stock II

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