動機

連續、小於等於最後一個值 就是用Monotonic Stack

Problem

Design an algorithm that 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 backward) for which the stock price 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].

Implement the StockSpanner class:

  • StockSpanner() Initializes the object of the class.
  • int next(int price) Returns the span of the stock's price given that today's price is price.

 

Example 1:

Input["StockSpanner", "next", "next", "next", "next", "next", "next", "next"][[], [100], [80], [60], [70], [60], [75], [85]]Output[Algorithm, Leetcode, 1, 1, 1, 2, 1, 4, 6]ExplanationStockSpanner stockSpanner = new StockSpanner();stockSpanner.next(100); // return 1stockSpanner.next(80);  // return 1stockSpanner.next(60);  // return 1stockSpanner.next(70);  // return 2stockSpanner.next(60);  // return 1stockSpanner.next(75);  // return 4, because the last 4 prices (including today's price of 75) were less than or equal to today's price.stockSpanner.next(85);  // return 6

 

Constraints:

  • 1 <= price <= 105
  • At most 104 calls will be made to next.

Sol

這是紀錄區間長度，，stack放(goal index, len)

class StockSpanner:

    def __init__(self):
        self.stk = [] # (now smallest price, bigger than this price cnt)

    def next(self, price: int) -> int:
        cnt = 1
        while self.stk and self.stk[-1][0] <= price: # 紀錄seg最大值
            cnt += self.stk.pop()[1]
        self.stk += [price, cnt],
        return cnt