leetCode练习（121）

题目：Best Time to Buy and Sell Stock

难度：easy

问题描述：

Say you have an array for which the i^th element is the price of a given stock on dayi.

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】//临时加的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.

解题思路：

很容易的一道题，遇到临时最小的，就保存（1），计算后面比它大的差，保留最大（6-1），遇到后面如果有更小的（0），就保存（0），重复上面工作即可。

代码如下：

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;    }