(LeetCode) 155. Min Stack

155. Min Stack

• Explore : Interview > Top Interveiw Questions > Easy Collection
• 분류 : Design
• 난이도 : Easy

Problem

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(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.

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 <= 2^31 - 1
• Methods pop, top and getMin operations will always be called on non-empty stacks.
• At most 3 * 10^4 calls will be made to push, pop, top, and getMin.

Solution

Exapnder

class MinStack() {

    /** initialize your data structure here. */
    var stack = Stack<Int>()

    fun push(`val`: Int) {
        stack.push(`val`)
    }

    fun pop() {
        stack.pop()
    }

    fun top(): Int {
        return stack.peek()
    }

    fun getMin(): Int {
        return stack.toList().min()!!
    }

}

/**
 * Your MinStack object will be instantiated and called as such:
 * var obj = MinStack()
 * obj.push(`val`)
 * obj.pop()
 * var param_3 = obj.top()
 * var param_4 = obj.getMin()
 */

Point of Thinking

• stack collection으로 편하게 구현했다.
• 디자인 조건의 top()은 peek()으로 대응하면 된다.
• getMin()은 list로 변환한 후 min()으로 추출하면 된다.
• min() 함수는 deprecated되어서 minOrNull()을 써야하지만, leetcode에서 지원하지 않으므로 min을 선택하였다.
• min()을 쓰면 Int?을 반환하여 !!를 사용했지만, 문제 조건상 getMin()호출시 stack은 비어있지않으므로 Accepted에는 지장이 없다.