Say you have an array 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 (ie, buy one and sell one share of the stock multiple times). However, you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
1:注意特殊情況;2:找到數組相鄰的凹點和凸點;3:兩者的差值是當前的最大值;4:在尋找凸凹值的時候注意邊界
int maxProfit(vector<int> &prices) { if(prices.size() <= 1) { return 0; } int maxValue = 0; int start = 0; int end = 0; int size = (int)prices.size(); while(start < size) { while(start < size - 1 && prices[start] >= prices[start + 1]) { start++; } end = start + 1; while(end < size - 1 && prices[end] <= prices[end + 1]) { end++; } if(end == size) { break; } else { maxValue += prices[end] - prices[start]; } start = end + 1; } return maxValue; }