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##### 题目描述：
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 (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.
Note that you cannot sell a stock before you buy one.
Input: [7,1,5,3,6,4] Output: 5 Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 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 Explanation: In this case, no transaction is done, i.e. max profit = 0.
##### 分析：
也就是找到非逆序元素差值的最大值，如果都是负值就返回0.
###### 方法一：
两轮循环，时间复杂度O(n^2)。
public int maxProfit(int[] prices) {if(prices==null|| prices.length==0)return 0;int res = 0;for(int i=0;i<prices.length;i++){for(int j=0;j<i;j++){if((prices[i]-prices[j])>res)res = prices[i] - prices[j];}}return res;}###### 方法二：
方法一复杂在于每次其实不需要求和之前所有元素的差值，只需要用一个变量记录之前最小的元素然后求差值即可。
public int maxProfit(int[] prices) {if(prices==null|| prices.length==0)return 0;int res = 0;int min=prices[0];for(int i=1;i<prices.length;i++){if(prices[i]<min)min = prices[i];else if((prices[i]- min)>res)res = prices[i] - min;}return res;}
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