題目：

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.

思路：

1.目標是求數組中的某兩個值的差（最大差值），前提是大的數在小的數后面

2.循環遍歷數組，設置一個當前最小值min(初始為Integer.MAX_VALUE)和一個最大差值max(初始化為0)

? ?每次更新min,并更新max(當前元素值-min)

public class Solution {public int maxProfit(int[] prices) {int len=prices.length;     //數組長度int max=0;                    //最大差值int min=Integer.MAX_VALUE;              //當前最小值for(int i=0;i<len;i++) {if(prices[i]<min)min=prices[i];if(prices[i]-min>max)max=prices[i]-min;}return max;}
}

　　

?

轉載于:https://www.cnblogs.com/hwu2014/p/4422009.html

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