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.

一刷
題解：在stack中存入的不是x, 而是x-min
例如3，4，2, 2
存入0，1（min=3), -1(min=2), 0
如果當前pop出來的小于0，則恢復原先的min

public class MinStack {
    Stack<Long> stack;
    long min = 0;

    /** initialize your data structure here. */
    public MinStack() {
        stack = new Stack<>();
    }
    
    public void push(int x) {
        if(stack.isEmpty()){
            stack.push(0L);
            min = x;
        }
        else{
            stack.push(x-min);
            if(x<min){
                min = x;
            }
        }
    }
    
    public void pop() {
        long cur = stack.pop();
        if(cur<0){
            long min_ori = min - cur;
            cur = cur + min_ori;
            min = min_ori;
        }
    }
    
    public int top() {
        long peek = stack.peek();
        if(peek>=0) return (int) (min+peek);
        else{
            long min_ori = min - peek;
            return (int) (peek + min_ori);
        }
    }
    
    public int getMin() {
        return (int)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();
 */