数据结构与算法

实现一个包含min函数的栈(O(1)的时间复杂度)

2019-12-21  本文已影响0人  而立之年的技术控
微信图片_20191221152107.jpg
class Solution:
    def __init__(self):
        self.stack = []
        self.min_stack = []
    def push(self, node):
        # write code here
        self.stack.append(node)
        if self.min_stack == []:
            self.min_stack.append(node)
        else:
            if node <= self.min_stack[-1]:
                self.min_stack.append(node)
            else:
                self.min_stack.append(self.min_stack[-1])
    def pop(self):
        # write code here
        if self.stack:
            self.min_stack.pop()
            return self.stack.pop()
    def top(self):
        # write code here
        if self.stack:
            return self.stack[-1]
    def min(self):
        # write code here
        if self.min_stack:
            return self.min_stack[-1]
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