[LeetCode] 901. 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.

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]。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/online-stock-span
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因为这道题的数据会到十的五次方，所以暴力解 N 方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。

注意如果问的这个跨度是不连续的，则无法用单调栈的思路做了。

时间O(n)

空间O(n)

Java实现

class StockSpanner {
    Stack<int[]> stack = new Stack<>();

    public StockSpanner() {
        
    }
    
    public int next(int price) {
        int res = 1;
        while (!stack.isEmpty() && stack.peek()[0] <= price) {
            res += stack.pop()[1];
        }
        stack.push(new int[]{price, res});
        return res;
    }
}

/**
 * Your StockSpanner object will be instantiated and called as such:
 * StockSpanner obj = new StockSpanner();
 * int param_1 = obj.next(price);
 */

 

单调栈题目汇总

LeetCode 题目总结