LeetCode【121】Best Time to Buy and Sell Stock

Say you have an array for which the ith element is the price of a given stock on day i.

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.

Example 1:

Input: [7, 1, 5, 3, 6, 4]Output: 5max. 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: 0In this case, no transaction is done, i.e. max profit = 0.

/** * 题意：最低价买进，最高价卖出，一次买入和卖出，求最大利润 * * @auther Dennis * @date 2017/12/13 * 思想：1、brute force蛮力法，两个for循环不断比较 *      2、双指针法，用一个min来保存买进的最低价，一个res来保存当天的价格减去最低价，也就是最大利润 */public class BestTimetoBuyandSellStock {    public static int maxProfit(int[] prices) {        int min = Integer.MAX_VALUE;        int res = 0;        for (int i = 0; i < prices.length; i++) {            if (prices[i] < min)                min = prices[i];            else if ((prices[i] - min) > res)                res = prices[i] - min;        }        return res;    }    public static void main(String[] args) {        int[] prices = new int[]{7, 1, 5, 3, 6};        int[] prices2 = new int[]{7, 6, 4, 3, 1};        System.out.println(maxProfit(prices));        System.out.println(maxProfit(prices2));    }}