題目：Best Time to Buy and Sell Stock

難度：easy

問題描述：

Say you have an array for which the i^th 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]【0 1 8】//臨時加的
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.

解題思路：

很容易的一道題，遇到臨時最小的，就儲存（1），計算後面比它大的差，保留最大（6-1），遇到後面如果有更小的（0），就儲存（0），重複上面工作即可。

程式碼如下：

public int maxProfit2(int[] prices) {
        if(prices==null||prices.length<=1){
        	return 0;
        }
        int res=0,temp,inv;
        temp=prices[0];
        for(int i=1;i<prices.length;i++){
        	inv=prices[i];
        	if(temp>inv){
        		temp=inv;
        	}else{
        		inv=inv-temp;
        		res=res>inv?res:inv;
        	}
        }
        return res;
    }