901.Online Stock Span

Tags: Medium Stack

Links: https://leetcode.com/problems/online-stock-span/

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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.

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单调栈的模型是找数组右边第一个大于当前值的下标（洛谷-P5788 [模板] 单调栈），虽然本题从求解目标上来说并不符合，但是通过在纸上模拟整个过程，发现可以将题目改写为从数组末尾看，左边第一个大于当前值的下标位置，所以栈内下标对应的实际数值应该是递减的。这里有个小细节，21行要先用res存储结果，因为如果没有这一行，最后返回n - 1 - s.top()会发现结果一直为0。

时间复杂度$O(n)$。

class StockSpanner {
    stack<int> s;
    vector<int> seq;
    int n;
public:
    StockSpanner() {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        n = 0;
    }

    int next(int price) {
        seq.push_back(price);
        ++n;

        while (!s.empty()) {
            if (seq[s.top()] <= price) s.pop();
            else {
                int res = n - 1 - s.top();
                s.push(n - 1);
                return res;
            }
        }

        s.push(n - 1); //推入下标
        return n;
    }
};

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