901. Online Stock Span

最新推荐文章于 2024-08-18 09:06:00 发布

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最新推荐文章于 2024-08-18 09:06:00 发布

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本文介绍了一个名为StockSpanner的类，该类收集股票每日价格并返回当前价格的跨度。跨度定义为从当前日期开始向后数最大连续天数，期间股票价格不大于当天的价格。文章通过一个实例展示了如何使用StockSpanner类进行操作。

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Write a class StockSpanner which collects daily price quotes for some stock, and returns the span of that stock's price for the current day.

The span of the stock's price today is defined as the maximum number of consecutive days (starting from today and going backwards) for which the price of the stock was less than or equal to today's price.

For example, if the price of a stock over the next 7 days were [100, 80, 60, 70, 60, 75, 85], then the stock spans would be [1, 1, 1, 2, 1, 4, 6].

Example 1:

Input: ["StockSpanner","next","next","next","next","next","next","next"], [[],[100],[80],[60],[70],[60],[75],[85]]
Output: [null,1,1,1,2,1,4,6]
Explanation: 
First, S = StockSpanner() is initialized.  Then:
S.next(100) is called and returns 1,
S.next(80) is called and returns 1,
S.next(60) is called and returns 1,
S.next(70) is called and returns 2,
S.next(60) is called and returns 1,
S.next(75) is called and returns 4,
S.next(85) is called and returns 6.

Note that (for example) S.next(75) returned 4, because the last 4 prices
(including today's price of 75) were less than or equal to today's price.

Note:

Calls to StockSpanner.next(int price) will have 1 <= price <= 10^5.
There will be at most 10000 calls to StockSpanner.next per test case.
There will be at most 150000 calls to StockSpanner.next across all test cases.
The total time limit for this problem has been reduced by 75% for C++, and 50% for all other languages.

维护一个单调递减序列，再维护一个数组保存加入的index顺序

注意结果是pop完自后末尾index+1

class StockSpanner(object):

    def __init__(self):
        self.a=[]
        self.idx=[]
        self.cnt=0
        
    def next(self, price):
        """
        :type price: int
        :rtype: int
        """
        self.cnt+=1
#        mi=self.cnt
        while self.a and self.a[-1]<=price: 
            self.a.pop()
            self.idx.pop()
        res=self.cnt-self.idx[-1] if self.idx else self.cnt
        self.a.append(price)
        self.idx.append(self.cnt)
        return res