"Leetcode" best time to Buy and Sell Stock II

Title:

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).

Tips:

This problem should use the idea of greedy algorithm. The greedy rules I use are as follows:

• Iterates from the first element to the last element;
• If the next element is larger than the current element, sell it, and buy at the price of one of the following elements, and accumulate the profit;
• If the next element is smaller than the current element, then the price of the following element is bought (if there is a continuous decrement, then the last buy price is the last of the smallest element);

Code:

classSolution { Public:    intMaxprofit (vector<int>&prices) {        if(prices.size () = =0)return 0; intbuy_in = prices[0], sum =0;  for(inti =1; I < prices.size (); ++i) {if(Prices[i] >buy_in) {Sum+ = Prices[i]-buy_in; Buy_in=Prices[i]; } Else{buy_in=Prices[i]; }        }        returnsum; }};

Later on the forum found a more concise solution, the core idea is actually more similar:

int maxprofit (vector<int> &prices) {    int0;      for 1; P < prices.size (); + +p       )10)        ; return ret;}

In fact, as long as there is a rise on the first sale and then buy, but this solution is more concise and clear.

"Leetcode" best time to Buy and Sell Stock II