package y2019.Algorithm.array;

/**

 * @ProjectName: cutter-point

 * @Package: y2019.Algorithm.array

 * @ClassName: MaxProfit2

 * @Author: xiaof

 * @Description: 122. Best Time to Buy and Sell Stock II

 * 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).

 *

 * 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.

 *

 * 你可以尽可能多的进行交易

 * 那么这个题就是有的赚就买的意思了

 *

 * @Date: 2019/7/2 10:12

 * @Version: 1.0

 */

public class MaxProfit2 {

    public int solution(int[] prices) {

        //那就是贪心算法了

        Integer curSmall = 0, result = 0;

        if(prices.length == 0)

            return 0; //没得赚

        else {

            curSmall = prices[0];

        }

        for(int i = 1; i < prices.length; ++i) {

            int temp = prices[i];

            if(curSmall == null || temp < curSmall) {

                curSmall = temp;

            } else if(temp > curSmall) {

                result += temp - curSmall;

                curSmall = temp;

            }

        }

        return result;

    }

    public static void main(String args[]) {

        int pres[] = {1,2,3,4,5};

        System.out.println(new MaxProfit2().solution(pres));

    }

}