LeetCode123 买卖股票的最佳时机 III(贪心)

LeetCode123 买卖股票的最佳时机 III

最多可以完成两笔交易,即分两段做LeetCode121,处理前缀后缀之后,遍历每一个分段节点所带来的收益取最大

class Solution:
    def maxProfit(self, prices: List[int]) -> int:

        pre_gains, suf_gains, pre_min, suf_max, l = [0], [0], prices[0], prices[-1], len(prices)
        
        cur_max = 0
        for i in range(l):
            pre_min = min(pre_min, prices[i])
            cur_max = max(cur_max, prices[i] - pre_min)
            pre_gains.append(cur_max)

        cur_max = 0
        for i in range(l - 1, -1, -1):
            suf_max = max(suf_max, prices[i])
            cur_max = max(cur_max, suf_max - prices[i])
            suf_gains.append(cur_max)

        ans = 0
        for i in range(l + 1):
            ans = max(ans, pre_gains[i] + suf_gains[l - i])

        return ans

posted on 2022-06-29 14:31  solvit  阅读(25)  评论(0编辑  收藏  举报

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