[leetcode] 739. Daily Temperatures @ python

最新推荐文章于 2023-12-21 23:27:29 发布

原创最新推荐文章于 2023-12-21 23:27:29 发布 · 568 阅读

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原题

Given a list of daily temperatures T, return a list such that, for each day in the input, tells you how many days you would have to wait until a warmer temperature. If there is no future day for which this is possible, put 0 instead.

For example, given the list of temperatures T = [73, 74, 75, 71, 69, 72, 76, 73], your output should be [1, 1, 4, 2, 1, 1, 0, 0].

Note: The length of temperatures will be in the range [1, 30000]. Each temperature will be an integer in the range [30, 100].

解法

堆栈. 构建stack储存还没有找到更大值的温度和索引, 初始化res为长度为len(T)值为0的列表, 遍历T, 如果走到i 点时发现i点的温度比stack中的温度要高, 则一直出栈, 更新res, 结果是两个index的差值.
Time: O(n)
Space: O(1)

代码

class Solution:
    def dailyTemperatures(self, T):
        """
        :type T: List[int]
        :rtype: List[int]
        """
        stack = []
        res = [0]*len(T)
        for i, temp in enumerate(T):
            while stack and stack[-1][1] < temp:
                prev_i = stack.pop()[0]
                res[prev_i] = i - prev_i
            stack.append([i, temp])
            
        return res