版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/cuicanxingchen123456/article/details/89036103
题目描述
假设您有一个数组,其中第i个元素是第一天给定股票的价格。如果你只被允许完成至多一笔交易(即买一份,卖一份股票),设计一个算法来找到最大利润。请注意,在购买股票之前,您不能出售股票。
可以有一次买入和一次卖出,那么买入必须在前。求最大收益。
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.
解题思路
使用贪心策略,假设第 i 轮进行卖出操作,买入操作价格应该在 i 之前并且价格最低。
public int maxProfit(int[] prices) {
if (prices == null || prices.length == 0)
return 0;
int soFarMin = prices[0];
int maxProfit = 0;
for (int i = 1; i < prices.length; i++) {
soFarMin = Math.min(soFarMin, prices[i]);
maxProfit = Math.max(maxProfit, prices[i] - soFarMin);
}
return maxProfit;
}