LeetCode 121. Best Time to Buy and Sell Stock--股票买入后再卖出，求最大收益，最多交易一次

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.

public class Main {    public int maxProfit(int[] prices) {        if(prices == null){            return 0;        }        int n = prices.length;        int minPrice = Integer.MAX_VALUE, maxProfit = 0;        for (int i = 0; i < n; i++) {            minPrice = Math.min(prices[i], minPrice);            maxProfit = Math.max(maxProfit, prices[i] - minPrice);        }        return maxProfit;    }    public static void main(String[] args) {        System.out.println(new Main().maxProfit(new int[]{7,1,5,3,6,4}));    }}//5

时间O(n)，空间O(1)。

200 / 200 test cases passed.
Status: Accepted
Runtime: 1 ms
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Your runtime beats 82.11 % of java submissions.