leetcode — best-time-to-buy-and-sell-stock-ii

/**
 * Source : https://oj.leetcode.com/problems/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 (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).
 */
public class BestTimeToBuyAndSellStock2 {

    /**
     * 找出最大可能的收益
     * 可以交易多次，但是同一时间手中只能有一只股票
     * 只要第二天的价格高于第一天，那么就可以在第一天买入，第二天卖出，这样就能保证每次赚钱，而且能赚到整个时间段内所有可能的利润
     *
     * @param prices
     * @return
     */
    public int maxProfit (int[] prices) {
        int result = 0;
        for (int i = 1; i < prices.length; i++) {
            result += prices[i] > prices[i-1] ? prices[i] - prices[i-1] : 0;
        }
        return result;
    }

    public static void main(String[] args) {
        BestTimeToBuyAndSellStock2 bestTimeToBuyAndSellStock2 = new BestTimeToBuyAndSellStock2();
        System.out.println(bestTimeToBuyAndSellStock2.maxProfit(new int[]{1,2,3,4,5}));
        System.out.println(bestTimeToBuyAndSellStock2.maxProfit(new int[]{-2,1,3,5,-9,1,2}));
    }
}

转载于:https://www.cnblogs.com/sunshine-2015/p/7837222.html