标签：remove etc append ret 进栈出栈 while amp may 一个
本题解法主要维护两个栈，一个是价格prices的单调递减栈。一个是weights栈随着prices 一起进栈出栈，计算w。
因为两个栈是一起进出，所以可以将两个栈写在一起Deque<List<>>
JAVA
class StockSpanner { Deque<Integer> prices; Deque<Integer> weights; public StockSpanner() { prices = new LinkedList<>(); weights = new LinkedList<>(); } public int next(int p) { int w = 1; while(!prices.isEmpty() && prices.getLast() <= p){ prices.removeLast(); w += weights.removeLast(); } prices.addLast(p); weights.addLast(w); return w; } }
class StockSpanner { Deque<List<Integer>> prices; public StockSpanner() { prices = new LinkedList<>(); } public int next(int p) { int w = 1; while(!prices.isEmpty() && prices.getLast().get(0) <= p){ w += prices.removeLast().get(1); } List<Integer> temp = new LinkedList<>(); temp.add(p); temp.add(w); prices.addLast(temp); return w; } }
Python3
class StockSpanner: def __init__(self): self.prices = [] def next(self, p): """ :type price: int :rtype: int """ w = 1 while len(self.prices) and self.prices[-1][0]<=p: w += self.prices.pop()[1] self.prices.append([p,w]) return w
May LeetCoding Challenge19 之 单调栈2.0
标签：remove etc append ret 进栈出栈 while amp may 一个
原文地址：https://www.cnblogs.com/yawenw/p/12943565.html
