[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 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.

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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思路

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。

这里补充一个为什么能想到单调栈里放的是{价钱，股票当日价格的跨度res}这个组合。比如题目给的例子[100, 80, 60, 70, 60, 75, 85]，一开始我们走到 100,80,60 都是递减的，所以直到 60，股价小于等于当前这一天的天数都是 0。然后我们走到70，此时我们发现70前面有一个60是小于70的，那么此时我就可以将60那天的结果（有多少天股价小于60）累加到70身上。因为股价小于60的天数一定也是小于等于70的。所以这道题的单调栈对于任何一天的股价而言，都是无条件入栈的，只是弹出的时候是有技巧的。当我们发现某一天 A 的股价大于栈顶那天 B 的股价的时候，就可以往外弹了，同时把所有小于 B 的股价的天数累加到 A 身上，意思是因为 B 天股价已经小于 A 天了，所以 如果有 X 天的股价是小于 B 那天的股价的，那么这些天的股价也必然小于 A 天。

复杂度

时间O(n) - worse case
空间O(n)

代码

Java实现

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

    public StockSpanner() {
        
    }
    
    public int next(int price) {
        int res = 1;
        while (!stack.isEmpty() && stack.peekLast()[0] <= price) {
            res += stack.pollLast()[1];
        }
        stack.offerLast(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);
 */