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).
我觉得这个就是贪心。对于prices[i] 往后找最后的prices[j] 然后把这段差值加上... 对于这种情况 1 5 5 10 当然取1买 10卖
class Solution { public: int maxProfit(vector<int>& prices) { int x = 0,ans = 0; int n = prices.size(); if (n <= 1) return 0; int be = 0; x = prices[be]; for (int i = 0; i < n; ++i) { if (prices[i] >= x) { x = prices[i]; } else { ans += x - prices[be]; be = i; if (be < n)x = prices[be]; } } if (x - prices[be] > 0) ans += x - prices[be]; return ans; } };