Say you have an array for which the i th 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.
(给定一个数组,它的第 i 个元素是一支给定股票第 i 天的价格。
如果你最多只允许完成一笔交易(即买入和卖出一支股票),设计一个算法来计算你所能获取的最大利润。
注意你不能在买入股票前卖出股票。)
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.
Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.
太简单了,遍历一边数组即可。Java实现代码如下:
class Solution {
public int maxProfit(int[] prices) {
if (prices == null || prices.length < 2) return 0;
int min = prices[0];
int profit = 0;
for (int price : prices) {
min = Math.min(min, price);
profit = Math.max(profit, price - min);
}
return profit;
}
}