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# 155 Min Stack

## 155. [Min Stack](https://leetcode.com/problems/min-stack/description/)

## 1. Question

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.
```

## 2. Implementation

思路：这题主要注意的是对于push()和pop()两种操作中，当我们发现min时该怎么办。当栈顶是min时，我们需要在pop了min后，可以立马知道当前min应该变为哪个数，所以我们的想法是每当我们发现当前的元素比min小时，我们立马将min压栈，然后更新min值。这样的好处是，当我们call pop()时，如果栈顶的元素等于min， 我们可以很快的知道min应该更新为哪个数

```java
class MinStack {
    int min;
    Stack<Integer> stack;

    /** initialize your data structure here. */
    public MinStack() {
        min = Integer.MAX_VALUE;
        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;
    }
}

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

## 3. Time & Space Complexity

所有操作的时间复杂度都是O(1)， 空间复杂度为O(n), n是minStack当前存储的元素个数