leecode100-买卖股票的最佳时期-贪心算法

MORE_772026-03-25 14:27

题目

解法1

思路：有题目可知，针对每一个price[i]，找到对于它来说，右边的最大值price[j]，获得max(price[j]-price[i])即可

复制代码

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        min_num = inf
        max_num = -1
        num = len(prices)
        max_list = [0] * num
        result = 0
        for i in range(num):
            j = num - i -1
            # print('j', j)
            max_num = max(prices[j], max_num)
            max_list[j] = max_num
        max_price = [max_list[i] - prices[i] for i in range(num)]
        result = 0 if max(max_price)<0 else max(max_price)
        return result

解法2

换个角度

思路：有题目可知，针对每一个price[i]，找到对于它来说，左边的最小值min_num，获得max(price[i]-min_num)即可

cpp 复制代码

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        num = len(prices)
        min_num = float('-inf')
        result = 0
        for i in range(num):
            min_num = min(min_num, prices[i])
            result = max(prices[i]-min_num, result)
        return result