Question

link

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.

You must implement a solution with O(1) time complexity for each function.

Analysis

辅助 stack （存最小值）。

I covered this question in another post [Question] Min Stack.

Code

class MinStack {

    Stack<Integer> stack = new Stack<Integer>();
    Stack<Integer> min = new Stack<Integer>();

    public void push(int x) {
        stack.push(x);
        if (min.isEmpty() || min.peek() >= x) {
            min.push(x);
        }
    }

    public void pop() {
        if (stack.isEmpty()) {
            return;
        }
        int topNum = stack.pop();
        if (topNum == min.peek()) {
            min.pop();
        }
    }

    public int top() {
        if (stack.isEmpty()) {
            return 0;
        }
        return stack.peek();
    }

    public int getMin() {
        if (min.isEmpty()) {
            return 0;
        }
        return min.peek();
    }
}