best-time-to-buy-and-sell-stock-ii.cpp

// 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 (i.e., buy one and sell one share of the stock multiple times).
//
// Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).
//
// Example 1:
//
//
// Input: [7,1,5,3,6,4]
// Output: 7
// Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
//              Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.
//
//
// Example 2:
//
//
// Input: [1,2,3,4,5]
// Output: 4
// Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
//              Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
//              engaging multiple transactions at the same time. You must sell before buying again.
//
//
// Example 3:
//
//
// Input: [7,6,4,3,1]
// Output: 0
// Explanation: In this case, no transaction is done, i.e. max profit = 0.
//


class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if (prices.size() <= 1) {
            return 0;
        }
        int max_profit = 0;
        auto size = prices.size();
        decltype(size) index = 1;
        while (index < size) {
            while (index<size && prices[index]<=prices[index-1]) {
                ++index;
            }
            if (index >= size) {
                return max_profit;
            }
            int min = prices[index-1];
            while (index<size && prices[index]>=prices[index-1]) {
                ++index;
            }
            if (index >= size) {
                return max_profit+prices[size-1]-min;
            }
            max_profit += prices[index-1]-min;
            ++index;
        }
        return max_profit;
    }
};