动态规划（1）

原题：

/** * Created by gouthamvidyapradhan on 17/03/2017. * Say you have an array for which the ith element is the price of a given stock on day i. * <p> * If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit. * <p> * Example 1: * Input: [7, 1, 5, 3, 6, 4] * Output: 5 * <p> * max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price) * Example 2: * Input: [7, 6, 4, 3, 1] * Output: 0 * <p> * In this case, no transaction is done, i.e. max profit = 0. */

答案：

public class BestTimeToBuyAndSellStocks {    /**     * Main method     *     * @param args     * @throws Exception     */    public static void main(String[] args) throws Exception {        int[] prices = {1, 1, 1, 1, 1};        System.out.println(new BestTimeToBuyAndSellStocks().maxProfit(prices));    }    public int maxProfit(int[] prices) {        if (prices.length == 0) return 0;        int[] max = new int[prices.length];        max[prices.length - 1] = prices[prices.length - 1];        for (int i = prices.length - 2; i >= 0; i--) {            max[i] = Math.max(prices[i], max[i + 1]);        }        int result = Integer.MIN_VALUE;        for (int i = 0, l = max.length; i < l; i++) {            result = Math.max(result, max[i] - prices[i]);        }        return result;    }}