翻译：假设你的数组中第i个元素是第i天的股票价格 。你最多只能做一次交易（买进并卖出一支股票，可以不做），设计一个算法来实现利润最大。
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: 5 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 In this case, no transaction is done, i.e. max profit = 0.
这个和找最大子数组的问题非常类似~个人认为关键点是卖出一定是在买入之后进行~
import java.util.Arrays;
import java.math.*;
class Solution {
public int maxProfit(int[] a) {
if (a.length==0)
return 0;
int profit = 0;//利润
int min=a[0];//最便宜的价格
for(int i=1;i<a.length;i++){
profit=Math.max(profit, (a[i]-min));//看之前的成交方式利润大，还是在本次卖出利润大
min=Math.min(min, a[i]);//看本次便宜还是之前的最低价便宜，相当于选最便宜的买入
}
return profit;
}
}