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  • 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}));
        }
    }
    
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  • 原文地址:https://www.cnblogs.com/sunshine-2015/p/7837222.html
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