##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.

Hint

Train of Thought

two stacks,一个用来存min值

one stack: 维护一个min值, 当新push的值小于min时,把min和新push的值都入栈, 然后更新min值 pop的时候检查是否等于min, 不等于就正常pop.如果等于min,说明它的下一个值就是第二小的,pop当前值之后, min=s.pop()

Code

public class MinStack {
    int min = Integer.MAX_VALUE;
    Stack<Integer> stack;
    /** initialize your data structure here. */
    public MinStack() {
        stack = new Stack<Integer>();
    }

    public void push(int x) {
        if (x <= min) {
            stack.push(min);
            min = x;
        }
        stack.push(x);
    }

    public void pop() {
        if (stack.pop() == min) {
            min = stack.pop();
        }
    }

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

    public int getMin() {
        return min;
    }
}

Complexity