155. Min Stack (E)

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

Implement the MinStack class:

• MinStack() initializes the stack object.

• void push(int val) pushes the element val onto the stack.

• void pop() removes the element on the top of the stack.

• int top() gets the top element of the stack.

• int getMin() retrieves the minimum element in the stack.

Example 1:

Input
["MinStack","push","push","push","getMin","pop","top","getMin"]
[[],[-2],[0],[-3],[],[],[],[]]

Output
[null,null,null,null,-3,null,0,-2]

Explanation
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); // return -3
minStack.pop();
minStack.top();    // return 0
minStack.getMin(); // return -2

Constraints:

• -231 <= val <= 231 - 1

• Methods pop, top and getMin operations will always be called on non-empty stacks.

• At most 3 * 104 calls will be made to push, pop, top, and getMin.

Solution:

使用两个仅支持 pop 和 push 的栈就可以完成, stack 储存压入的数据, minStack 储存最小值.

• push 直接把元素压入 stack, 对于 minStack, 如果它为空则直接压入, 反之压入当前元素与 minStack 栈顶的最小值

• pop 两个栈都弹出一个元素, 返回 stack 弹出的元素

• min 返回 minStack 的栈顶

还可以令 minStack 为单调栈, 即push时只有元素更小的时候才放入这个栈, 而pop时只有栈顶与stack栈顶相同时才弹出

这样可以节约一定的空间, 但是实质上空间复杂度仍然是 O(n), 且多了一些判断, 并不一定更优

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

    public MinStack() {
        stack = new Stack<Integer>(); 
        minStack = new Stack<Integer>();
    }
    
    public void push(int val) {
        stack.push(val);
        
        if(minStack.isEmpty())
        {
            minStack.push(val);
        }
        else
        {
            minStack.push(Math.min(minStack.peek(), val));
        }
        
    }
    
    public void pop() {
        stack.pop();
        minStack.pop();
        
    }
    
    public int top() {
        return stack.peek();
        
    }
    
    public int getMin() {
        return minStack.peek();
        
    }
}

Version 2: Mono stack

public MinStack() {
    stack = new Stack<Integer>();
    minStack = new Stack<Integer>();
}

public void push(int number) {
    stack.push(number);
    if (minStack.empty() == true)
        minStack.push(number);
    else if (minStack.peek() >= number) // 这里考虑的相等的情况也会继续push
        minStack.push(number);
}

public int pop() {
    if (stack.peek().equals(minStack.peek()))
        minStack.pop();
    return stack.pop();
}

public int min() {
    return minStack.peek();