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901. Online Stock Span ​

题目 ​

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 in one day is the maximum number of consecutive days (starting from that day and going backward) for which the stock price was less than or equal to the price of that day.

  • For example, if the prices of the stock in the last four days is [7,2,1,2] and the price of the stock today is 2, then the span of today is 4 because starting from today, the price of the stock was less than or equal 2 for 4 consecutive days.
  • Also, if the prices of the stock in the last four days is [7,34,1,2] and the price of the stock today is 8, then the span of today is 3 because starting from today, the price of the stock was less than or equal 8 for 3 consecutive days.

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.
Related Topics
  • 栈
  • 设计
  • 数据流
  • 单调栈

  • 👍 501
  • 👎 0

    • 思路 ​

    本地需要处理新增处理重复的元素，如果重复的元素也存在栈内，直接弹出，保留最后入栈的元素

    解法 ​

    py
    # leetcode submit region begin(Prohibit modification and deletion)
class StockSpanner:

    def __init__(self):
        self.stack = []
        self.list = []

    def next(self, price: int) -> int:
        self.list.append(price)
        size = len(self.list)
        while self.stack and price >= self.list[self.stack[-1]]:
            self.stack.pop()
        cur_index = size - 1
        short_after_max = 0 if not self.stack else self.stack[-1] + 1

        res = cur_index - short_after_max
        self.stack.append(cur_index)
        return res + 1




        


# Your StockSpanner object will be instantiated and called as such:
# obj = StockSpanner()
# param_1 = obj.next(price)
# leetcode submit region end(Prohibit modification and deletion)

    复杂度分析 ​

    • 时间复杂度 O(N)
    • 空间复杂度 O(N)