155. Min Stack

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.

//Approach 1: using two stack, Time: O(n), Space: O(n)
class MinStack {
    Stack<Integer> s;
    Stack<Integer> m;
    /** initialize your data structure here. */
    public MinStack() {
        s = new Stack<Integer>();
        m = new Stack<Integer>();
    }
    
    public void push(int x) {
        s.push(x);
        
        if (m.isEmpty() || x <= m.peek()) {//不要忘记判断m是否为空
            m.push(x);
        }
    }
    
    public void pop() {
        int p = s.pop();
        
        if (!m.isEmpty() && p == m.peek()) {
            m.pop();
        }
    }
    
    public int top() {
        return s.peek();
    }
    
    public int getMin() {
        return m.peek();
    }
}

//Approach 2: using one stack, Time: O(n), Space: O(n)
//思路就是记录一个global的min，每次当x小于min时，先把min入栈然后x入栈，这样当出栈的是min时，在pop一次就相当于找到之前的min
class MinStack {
    Stack<Integer> s;
    int min = Integer.MAX_VALUE;
    /** initialize your data structure here. */
    public MinStack() {
        s = new Stack<Integer>();
    }
    
    public void push(int x) {
        if (x <= min) {
            s.push(min);
            min = x;
        }
        s.push(x);
    }
    
    public void pop() {
        if (s.pop() == min) {
           min = s.pop();
        }
    }
    
    public int top() {
        return s.peek();
    }
    
    public int getMin() {
        return min;
    }
}