Description

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

  • push(x) – Push element x onto stack.
  • pop() – Removes the element on top of the stack.
  • top() – Get the top element.
  • getMin() – Retrieve the minimum element in the stack.

Example:

MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin();   --> Returns -3.
minStack.pop();
minStack.top();      --> Returns 0.
minStack.getMin();   --> Returns -2.

Solutions

1. 两个 Stack

class MinStack:

    def __init__(self):
        """
        initialize your data structure here.
        """
        self.stack = []
        self.min = []

    def push(self, x: int) -> None:
        if len(self.min) == 0 or x <= self.min[-1]:
            self.min.append(x)
        self.stack.append(x)

    def pop(self) -> None:
        if self.stack[-1] == self.min[-1]:
            self.min.pop()
        self.stack.pop()

    def top(self) -> int:
        return self.stack[-1]

    def getMin(self) -> int:
        return self.min[-1]
# Runtime: 68 ms, faster than 91.20% of Python3 online submissions for Min Stack.
# Memory Usage: 17.5 MB, less than 5.36% of Python3 online submissions for Min Stack.

# Your MinStack object will be instantiated and called as such:
# obj = MinStack()
# obj.push(x)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.getMin()

References

  1. 155. Min Stack