LintCode:买卖股票的最佳时机 III

最新推荐文章于 2025-03-22 08:09:54 发布

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最新推荐文章于 2025-03-22 08:09:54 发布

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LintCode:买卖股票的最佳时机 III

直接暴力拆分成一前一后两个数组，分别求解，可惜超时，暂时还没想到更好的解法，容我再想想。

class Solution:
    """
    @param prices: Given an integer array
    @return: Maximum profit
    """
    def maxProfit(self, prices):
        # write your code here
        if not prices or len(prices) == 1:            
            return 0      
        n = len(prices)
        ans = 0
        for i in range(n+1):
            pre_prices = prices[0:i]
            last_prices = prices[i:]
            pre_ans = 0
            last_ans = 0
            if pre_prices:
                pre_lowest = pre_prices[0]
            if last_prices:
                last_lowest = last_prices[0]
            for j in range(len(pre_prices)):
                pre_ans = max(pre_ans, pre_prices[j] - pre_lowest)
                pre_lowest = min(pre_lowest, pre_prices[j])
            for j in range(len(last_prices)):
                last_ans = max(last_ans, last_prices[j] - last_lowest)
                last_lowest = min(last_lowest, last_prices[j])
            ans = max(pre_ans + last_ans, ans)
        return ans