LC.155. Min Stack(非优化，两个stack 同步 ＋ －)

https://leetcode.com/problems/min-stack/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.

corner cases:

1)If the stack is empty, min() should return -1.

2)pop() - remove and return the top element, if the stack is empty, return -1

3)push(int element) - push the element to top
top() - return the top element without remove it, if the stack is empty, return -1

4) min() - return the current min value in the stack.

这种写法 stack 1 stack 2 同步操作，stack 1 进来一个  stack2 也要进来一个。如果有特别多重复的数进来的情况，并且是一整块一样的数进来，那是可以被优化的。

1111111  222222  -1-1-1-1 111111 -2-2-2-2-2 33333 

如果 进来的数字是交叉的，则优化空间有限: 1212121212

public class LC_155_MinStack {

    private Deque<Integer> stack ;
    private Deque<Integer> minStack ;

    /** initialize your data structure here. */
    public LC_155_MinStack() {
        stack = new LinkedList<>() ;
        minStack = new LinkedList<>();
    }
    /*
     stack -2  0   -3
     min   -2  -2  -3
    * */
    public void push(int x) {
        stack.push(x);
        if (minStack.isEmpty()){
            minStack.push(x);
        } else{
            if (minStack.peek() < x){
                minStack.push(minStack.peek());
            } else{
                minStack.push(x);
            }
        }
    }

    public void pop() {
        if (stack.isEmpty()) return ;
        stack.pop();
        minStack.pop();
    }
    //peek
    public int top() {
        if (stack.isEmpty()) return -1 ;
        return stack.peek() ;
    }

    public int getMin() {
        if (minStack.isEmpty()) return -1 ;
        return minStack.peek();
    }
}