leetcode：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.

其实就是求前后两个数最大的差值，当然前提是前面的数更小。

我们可以遍历一次这个数组，在遍历的过程中记录下当前的最小值，并且计算出在当前点可以获得的最大值  整个算法的复杂度O（N）

public class Solution {    public int maxProfit(int[] prices)     {        int n=prices.length;        int s=0;        if(n==0)            return 0;        int minNum=prices[0];        for(int i=1;i<n;i++)        {            if(prices[i]<minNum)                minNum=prices[i];            else if(prices[i]-minNum>s)                s=prices[i]-minNum;        }        return s;    }}