【leetcode】121. Best Time to Buy and Sell Stock

题目要求:

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.

即选一天买，选一天卖，使得收益最大化

刚开始的思路：从第一个价格开始，和后面每一天的价格相减，得到一个为负的最小值（因为买入价低于卖出价），转换为整数即为最大收益。但是超出了时间限制。

public int maxProfit(int[] prices) {        int min = 0;        for(int i=0;i<prices.length;i++)        {            for(int j=i;j<prices.length;j++)            {                if(prices[i]-prices[j]<min)                {                    min=prices[i]-prices[j];                }            }        }        int max = 0-min;        return max;    }

看了网上其他思路，更改代码如下：

public class Solution {    public int maxProfit(int[] prices) {          if(prices==null||prices.length<=1)    {        return 0;    }    int min = prices[0];    int result = 0;    for(int i=0;i<prices.length;i++)    {        result=Math.max(result,prices[i]-min);        min = Math.min(min,prices[i]);    }    return result;    }}