Say you have an array for which theithelement 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]
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.

分析：

在一个array，找一个最低点和一个最高点，且满足最低点在前，最高点在后。

int minValuel; 遍历arr，如果arr[i]< minValue, minValue = arr[i];

int maxProfit; 如果arr[i]-minVal >maxprofit，maxprofit= arr[i]-minVal

public class Solution {
    public int maxProfit(int[] prices) {

    if(prices == null || prices.length <= 1){
        return  0;
    }  
    int maxProfit = 0;
    int minValue = Integer.MAX_VALUE;    
    for(int i = 0; i < prices.length; i++){
        if(prices[i] < minValue){
            minValue = prices[i];
            continue;
        }

        if(prices[i] - minValue > maxProfit){
            maxProfit = prices[i] - minValue;
        }
    }    
    return maxProfit;
    }
}