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87 lines
2.0 KiB
Python
87 lines
2.0 KiB
Python
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155. Min Stack
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Difficulty: Medium
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https://leetcode.com/problems/min-stack/
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──────────────────────────────────────────────────
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Design a stack that supports push, pop, top, and retrieving the
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minimum element in constant time.
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Implement the MinStack class:
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• MinStack() initializes the stack object.
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• void push(int value) pushes the element value onto the stack.
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• void pop() removes the element on the top of the stack.
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• int top() gets the top element of the stack.
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• int getMin() retrieves the minimum element in the stack.
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You must implement a solution with O(1) time complexity for each
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function.
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Example 1:
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Input
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["MinStack","push","push","push","getMin","pop","top","getMin"]
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[[],[-2],[0],[-3],[],[],[],[]]
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Output
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[null,null,null,null,-3,null,0,-2]
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Explanation
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MinStack minStack = new MinStack();
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minStack.push(-2);
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minStack.push(0);
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minStack.push(-3);
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minStack.getMin(); // return -3
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minStack.pop();
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minStack.top(); // return 0
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minStack.getMin(); // return -2
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Constraints:
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• -2^31 <= val <= 2^31 - 1
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• Methods pop, top and getMin operations will always be called on
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non-empty stacks.
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• At most 3 * 10^4 calls will be made to push, pop, top, and getMin.
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"""
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class MinStack:
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def __init__(self):
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self.stack = []
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self.minStack = []
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def push(self, value: int) -> None:
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self.stack.append(value)
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value = min(value, self.minStack[-1] if self.minStack else value)
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self.minStack.append(value)
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def pop(self) -> None:
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self.stack.pop()
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self.minStack.pop()
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def top(self) -> int:
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return self.stack[-1]
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def getMin(self) -> int:
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return self.minStack[-1]
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# Your MinStack object will be instantiated and called as such:
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# obj = MinStack()
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# obj.push(value)
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# obj.pop()
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# param_3 = obj.top()
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# param_4 = obj.getMin()
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