# 买卖股票的最佳时机 Best Time to Buy and Sell Stock

## 题目

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 (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Note that you cannot sell a stock before you buy one.

Example 1:

``````Input: [7,1,5,3,6,4]
Output: 5
``````

Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 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
``````

Explanation: In this case, no transaction is done, i.e. max profit = 0.

## 思路&题解

1. 每一次买入之后，剩下的时间循环计算所有卖出可能性，算出最大利润；
2. 冲第一天开始，循环计算买入价格，结合步骤一得到每次不同买入后可能得到的最大利润，取大值。
``````public int maxProfit(int[] prices) {
int maxProfit=0;
for(int i = 0; i < prices.length; i++){
int buy = prices[i];
int sell = buy;
for(int j = i+1; j < prices.length; j++){
if(prices[j] > sell) {
sell = prices[j];
}
}
maxProfit = Math.max(sell-buy, maxProfit);
}
return maxProfit;
}
``````

## 知识点分析

``````public class Solution {
public int maxProfit(int prices[]) {
int minprice = Integer.MAX_VALUE;
int maxprofit = 0;
for (int i = 0; i < prices.length; i++) {
if (prices[i] < minprice)
minprice = prices[i];
else if (prices[i] - minprice > maxprofit)
maxprofit = prices[i] - minprice;
}
return maxprofit;
}
}
``````