901. Online Stock Span
- Hardness: \(\color{orange}\textsf{Medium}\)
- Ralated Topics:
Stack、Design、Monotonic Stack、Data Stream
一、題目
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 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 theStockSpannerclass: -
StockSpanner()Initializes the object of the class. -
int next(int price)Returns the span of the stock’s price given that today’s price isprice.
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 <= 10^5- At most
10^4calls will be made tonext.
二、分析
- 可以用
Monotonic Stack裝載pair<int,int>記錄當前stock的股價與天數，假如遇到當前股價還低者，便pop掉，當前股價與Stack頂的位置差即為低於當前股價的連續天數。
三、解題
1. Monotonic Stack
- Time complexity: \(O(n)\)
- Space complexity: \(O(n)\)
class StockSpanner {
public:
stack<pair<int,int>> st;
int day = 0; // 記錄第 ith day
StockSpanner() {
st.push(make_pair(INT_MAX, -1)); // 令 stack 不會為空
}
int next(int price) {
while (st.top().first <= price) st.pop(); // 遇到小於等於的都 pop掉
int res = day - st.top().second; // 堆頂的位置與當前位置差即為解
st.push(make_pair(price, day++)); // 再將當前股價推到堆中
return res;
}
};