알고리즘 문제/Leetcode
155. Min Stack
BEstyle
2022. 12. 28. 01:24
Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.
Implement the MinStack class:
- MinStack() initializes the stack object.
- void push(int val) pushes the element val onto the stack.
- void pop() removes the element on the top of the stack.
- int top() gets the top element of the stack.
- int getMin() retrieves the minimum element in the stack.
You must implement a solution with O(1) time complexity for each function.
Example 1:
Input
["MinStack","push","push","push","getMin","pop","top","getMin"]
[[],[-2],[0],[-3],[],[],[],[]]
Output
[null,null,null,null,-3,null,0,-2]
Explanation
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); // return -3
minStack.pop();
minStack.top(); // return 0
minStack.getMin(); // return -2
Constraints:
- -231 <= val <= 231 - 1
- Methods pop, top and getMin operations will always be called on non-empty stacks.
- At most 3 * 104 calls will be made to push, pop, top, and getMin.
class MinStack:
def __init__(self):
self.alist=[]
def push(self, val: int) -> None:
self.alist.append(val)
def pop(self) -> None:
self.alist.pop()
def top(self) -> int:
return self.alist[-1]
def getMin(self) -> int:
return min(self.alist)
# Your MinStack object will be instantiated and called as such:
# obj = MinStack()
# obj.push(val)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.getMin()