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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 have1 <= price <= 10^5
. - There will be at most
10000
calls toStockSpanner.next
per test case. - There will be at most
150000
calls toStockSpanner.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