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(biāo)是求數(shù)組中的某兩個(gè)值的差（最大差值），前提是大的數(shù)在小的數(shù)后面

2.循環(huán)遍歷數(shù)組，設(shè)置一個(gè)當(dāng)前最小值min(初始為Integer.MAX_VALUE)和一個(gè)最大差值max(初始化為0)

? ?每次更新min,并更新max(當(dāng)前元素值-min)

public class Solution {public int maxProfit(int[] prices) {int len=prices.length;     //數(shù)組長(zhǎng)度int max=0;                    //最大差值int min=Integer.MAX_VALUE;              //當(dāng)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(zhuǎn)載于:https://www.cnblogs.com/hwu2014/p/4422009.html

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