Python 1 line of code for each definition and only using 1 stack (feedback welcome)

  • -1

    The code is slightly sped up by computing self.getMin() once in 'push' vs. 3 times.

    class MinStack(object):
        def __init__(self):
            self.stack = []
        def push(self, x):
            self.stack.append((x, x if self.getMin() is None or x < self.getMin() else self.getMin()))
        def pop(self):
        def top(self):
            return None if not self.stack else self.stack[-1][0]
        def getMin(self):
            return None if not self.stack else self.stack[-1][1]

  • 0

    these down votes without any guidance for improvement are quite discouraging :(

  • 0

    = = this is essentially the same with two stacks, man.
    but it consumes more memory.

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