[LeetCode] 901. Online Stock Span 线上股票跨度

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:

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

给一个股票价格的数组，写一个函数计算每一天的股票跨度，股票跨度是从当天股票价格开始回看连续的比当天价格低的最大连续天数。

解法1：简单但效率低的思路，对于每一天向后计算连续比当前价格低的天数。时间复杂度：O(n^2)

解法2: 栈

You can refer to the same problem 739. Daily Temperatures.

Push every pair of <price, result> to a stack.
Pop lower price from the stack and accumulate the count.

One price will be pushed once and popped once.
So 2 * N times stack operations and N times calls.
I'll say time complexity is O(1)

Java:

Stack<int[]> stack = new Stack<>();
    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;
    }　　

Python:

def __init__(self):
        self.stack = []
 
    def next(self, price):
        res = 1
        while self.stack and self.stack[-1][0] <= price:
            res += self.stack.pop()[1]
        self.stack.append([price, res])
        return res

C++:

stack<pair<int, int>> s;
    int next(int price) {
        int res = 1;
        while (!s.empty() && s.top().first <= price) {
            res += s.top().second;
            s.pop();
        }
        s.push({price, res});
        return res;
    }