1986

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.

Method 1. two stacks 

class MinStack {
    private Stack<Integer> stack;
    private Stack<Integer> minstack;

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

    public void push(int x) {
        stack.push(x);
        if(minstack.empty()) {
            minstack.push(x);
        }else {
            minstack.push(Math.min(x, minstack.peek()));
        }
    }

    public void pop() {
        if(stack.empty()) return;
        stack.pop();
        minstack.pop();
    }

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

    public int getMin() {
        return minstack.peek();
    }
}


Method 2. one stack (每次遇见最小值时候要把之前那个最小值也存一下，要不最小值pop掉之后就找不到第二个最小值了)

class MinStack {
    Stack<Integer> stack;
    int min = Integer.MAX_VALUE;

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

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

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

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

    public int getMin() {
        return min;
    }
}