[Swift]LeetCode901. 股票价格跨度 | Online Stock Span
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?微信公眾號:山青詠芝(shanqingyongzhi)
?博客園地址:山青詠芝(https://www.cnblogs.com/strengthen/)
?GitHub地址:https://github.com/strengthen/LeetCode
?原文地址:?https://www.cnblogs.com/strengthen/p/10607919.html?
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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:
- 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.
編寫一個?StockSpanner?類,它收集某些股票的每日報價,并返回該股票當日價格的跨度。
今天股票價格的跨度被定義為股票價格小于或等于今天價格的最大連續日數(從今天開始往回數,包括今天)。
例如,如果未來7天股票的價格是?[100, 80, 60, 70, 60, 75, 85],那么股票跨度將是?[1, 1, 1, 2, 1, 4, 6]。?
示例:
輸入:["StockSpanner","next","next","next","next","next","next","next"], [[],[100],[80],[60],[70],[60],[75],[85]] 輸出:[null,1,1,1,2,1,4,6] 解釋: 首先,初始化 S = StockSpanner(),然后: S.next(100) 被調用并返回 1, S.next(80) 被調用并返回 1, S.next(60) 被調用并返回 1, S.next(70) 被調用并返回 2, S.next(60) 被調用并返回 1, S.next(75) 被調用并返回 4, S.next(85) 被調用并返回 6。注意 (例如) S.next(75) 返回 4,因為截至今天的最后 4 個價格 (包括今天的價格 75) 小于或等于今天的價格。?
提示:
- 調用?
StockSpanner.next(int price)?時,將有?1 <= price <= 10^5。 - 每個測試用例最多可以調用??
10000?次?StockSpanner.next。 - 在所有測試用例中,最多調用?
150000?次?StockSpanner.next。 - 此問題的總時間限制減少了 50%。
Runtime:?840 ms Memory Usage:?23 MB
1 class StockSpanner { 2 var spans:[Int] 3 var prices:[Int] 4 var index:Int 5 6 init() { 7 spans = [Int](repeating:0,count:10_000) 8 prices = [Int](repeating:0,count:10000) 9 index = -1 10 } 11 12 func next(_ price: Int) -> Int { 13 index += 1 14 prices[index] = price 15 if index == 0 || price < prices[index - 1] 16 { 17 spans[index] = 1 18 return 1 19 } 20 var previousIndex:Int = index - 1 21 var span:Int = 1 22 while (previousIndex >= 0 && price >= prices[previousIndex]) 23 { 24 span += spans[previousIndex] 25 previousIndex -= spans[previousIndex] 26 } 27 spans[index] = span 28 return span 29 } 30 } 31 32 /** 33 * Your StockSpanner object will be instantiated and called as such: 34 * let obj = StockSpanner() 35 * let ret_1: Int = obj.next(price) 36 */
892ms
1 class StockSpanner { 2 3 private var s = [(Int, Int)]() 4 init() { 5 6 } 7 8 func next(_ price: Int) -> Int { 9 var sum = 1 10 while !s.isEmpty, s.last!.0 <= price { 11 sum += s.removeLast().1 12 } 13 s.append((price, sum)) 14 return sum 15 } 16 }
928ms
1 class StockSpanner { 2 3 init() { 4 5 } 6 7 struct PriceSpan { 8 let price: Int 9 let span: Int 10 } 11 12 var stack = [PriceSpan]() 13 14 func next(_ price: Int) -> Int { 15 guard stack.count > 0 else { 16 stack.append(PriceSpan(price: price, span: 1)) 17 return 1 18 } 19 20 var span = 1 21 while stack.last != nil && stack.last!.price <= price { 22 span += stack.last!.span 23 stack.removeLast() 24 } 25 26 stack.append(PriceSpan(price: price, span: span)) 27 return span 28 } 29 } 30 31 /** 32 * Your StockSpanner object will be instantiated and called as such: 33 * let obj = StockSpanner() 34 * let ret_1: Int = obj.next(price) 35 */
1036ms
1 class StockSpanner { 2 private var span: [Int] = [] 3 private var stack: [StockSpan] = [] 4 init() { 5 6 } 7 8 func next(_ price: Int) -> Int { 9 var stockSpan = StockSpan(price:price, span: 1) 10 while !stack.isEmpty && stack.last!.price <= stockSpan.price { 11 let removed = stack.removeLast() 12 stockSpan.span += removed.span 13 } 14 stack.append(stockSpan) 15 return stockSpan.span 16 } 17 18 struct StockSpan { 19 let price: Int 20 var span: Int 21 } 22 } 23 24 /** 25 * Your StockSpanner object will be instantiated and called as such: 26 * let obj = StockSpanner() 27 * let ret_1: Int = obj.next(price) 28 */
20764 kb
1 class StockSpanner { 2 var prices: [Int] = [] 3 var days: [Int] = [] 4 init() { 5 6 } 7 8 func next(_ price: Int) -> Int { 9 if prices.isEmpty || prices.last! > price { 10 prices.append(price) 11 days.append(1) 12 return 1 13 } 14 var index = prices.count - 1 15 var res = 1 16 while index >= 0 && prices[index] <= price { 17 res += days[index] 18 index -= days[index] 19 } 20 prices.append(price) 21 days.append(res) 22 return res 23 } 24 } 25 26 /** 27 * Your StockSpanner object will be instantiated and called as such: 28 * let obj = StockSpanner() 29 * let ret_1: Int = obj.next(price) 30 */
20820 kb
1 class StockSpanner { 2 3 private var elements : [(price : Int, conquered : Int)] = [] 4 init() { 5 6 } 7 8 func next(_ price: Int) -> Int { 9 var conquered : Int = 1 10 while !elements.isEmpty && elements.last!.price <= price { 11 let removed = elements.removeLast() 12 conquered += removed.conquered 13 } 14 15 elements.append((price: price, conquered: conquered)) 16 return elements.last!.conquered 17 } 18 } 19 20 /** 21 * Your StockSpanner object will be instantiated and called as such: 22 * let obj = StockSpanner() 23 * let ret_1: Int = obj.next(price) 24 */
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轉載于:https://www.cnblogs.com/strengthen/p/10607919.html
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