901. Online Stock Span

https://leetcode.com/problems/online-stock-span/

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
[null, 1, 1, 1, 2, 1, 4, 6]

Explanation
StockSpanner stockSpanner = new StockSpanner();
stockSpanner.next(100); // return 1
stockSpanner.next(80);  // return 1
stockSpanner.next(60);  // return 1
stockSpanner.next(70);  // return 2
stockSpanner.next(60);  // return 1
stockSpanner.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.

实时每次进来一个数都统计前面连续等于或小于当前数的元素个数。暴力法O(N^2)，连续数组subarray => DP, 窗口或presum。dp[i]记录当前元素时连续最长长度，i的price如果>=dp[i-1] => dp[i] += dp[i-1]，此时还没有结束，因为price[i - dp[i - 1]]也可能比price小，所以i继续跳转到i - dp[i-1]，以此类推直到不满足>=为止。进一步可把中间用于跳转的状态全部合并，因为subarray的性质如果跳转的前一步条件不满足，后面的也就无需跳转，中间的跳转状态也就无需存在，具体可参考花花的图：

class StockSpanner:

    def __init__(self):
        self.s = []

    def next(self, price: int) -> int:
        res = 1
        while self.s and price >= self.s[-1][0]:
            res += self.s[-1][1]
            self.s.pop()
        self.s.append((price, res))
        return res


# Your StockSpanner object will be instantiated and called as such:
# obj = StockSpanner()
# param_1 = obj.next(price)