[LeetCode] 901. Online Stock Span

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:

1. Calls to StockSpanner.next(int price) will have 1 <= price <= 10^5.
2. There will be at most 10000 calls to StockSpanner.next per test case.
3. There will be at most 150000 calls to StockSpanner.next across all test cases.
4. The total time limit for this problem has been reduced by 75% for C++, and 50% for all other languages.

股票价格跨度。编写一个 StockSpanner 类，它收集某些股票的每日报价，并返回该股票当日价格的跨度。今天股票价格的跨度被定义为股票价格小于或等于今天价格的最大连续日数（从今天开始往回数，包括今天）。例如，如果未来7天股票的价格是 [100, 80, 60, 70, 60, 75, 85]，那么股票跨度将是 [1, 1, 1, 2, 1, 4, 6]。

因为这道题的数据会到十的五次方，所以N平方时间复杂度的思路应该是会超时的。O(n)级别的思路是用单调栈。具体实现是用单调栈放两个东西，{价钱，股票当日价格的跨度res}。遍历input的数组，当stack为空的时候，就直接入栈；当stack不为空，需要查看栈顶元素是否小于等于当前要入栈的price，如果是的话就弹出栈顶元素，并把栈顶元素的res累加到当前要入栈的元素的res里去。

时间O(n)

空间O(n)

Java实现

 1 class StockSpanner {
 2     Stack<int[]> stack = new Stack<>();
 3 
 4     public StockSpanner() {
 5         
 6     }
 7     
 8     public int next(int price) {
 9         int res = 1;
10         while (!stack.isEmpty() && stack.peek()[0] <= price) {
11             res += stack.pop()[1];
12         }
13         stack.push(new int[]{price, res});
14         return res;
15     }
16 }
17 
18 /**
19  * Your StockSpanner object will be instantiated and called as such:
20  * StockSpanner obj = new StockSpanner();
21  * int param_1 = obj.next(price);
22  */

单调栈题目汇总

LeetCode 题目总结