BestTimetoBuyandSellStockII.java

package easy;
/**
 * 
 * ClassName: BestTimetoBuyandSellStockII
 * @author chenyiAlone  
 * Create Time: 2019/03/08 17:38:03
 * Description: No.122
 * 思路:
 *     因为可以进行多次的买卖，那么只需要贪心得在两个相邻的进行判断是否赚钱，将所有赚的钱加起来就是最大利润
 * 
 * 
 * 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.
 */
public class BestTimetoBuyandSellStockII {
    public int maxProfit(int[] prices) {
        int res = 0;
        for (int i = 1; i < prices.length; i++) {
            int profit = prices[i] - prices[i - 1];
            if (profit > 0) 
                res += profit;
        }
        return res;
    }
}