题解 | #牛牛的奶牛买卖#
牛牛的奶牛买卖
https://www.nowcoder.com/practice/d32969dc6a5c4f70a00a02b642a4f4cd?tpId=363&tqId=10625843&ru=/exam/oj&qru=/ta/super-company23Year/question-ranking&sourceUrl=%2Fexam%2Foj%3Fpage%3D1%26tab%3D%25E7%25AE%2597%25E6%25B3%2595%25E7%25AF%2587%26topicId%3D363
import java.util.*;
public class Solution {
/**
* 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
*
*
* @param prices int整型一维数组
* @return int整型
*/
public int maxProfit (int[] prices) {
int minPrice = Integer.MAX_VALUE;
int maxProfit = 0;
for (int i = 0; i < prices.length; i++) {
minPrice = Math.min(minPrice, prices[i]);
maxProfit = Math.max(maxProfit, prices[i] - minPrice);
}
return maxProfit;
}
}
本题知识点分析:
1.动态规划
2.API函数(Math.max,Math.min)
3.数学模拟
4.数组遍历
本题解题思路分析:
1.一边遍历一边记录最小值
2.每次元素去和最小值做差值比较,找到最大的差值返回

