leetcode155.最小栈

 思路源自

【力扣hot100】【LeetCode 155】最小栈

为了让检索时间达到o（1），采用空间换时间，维护两个栈，第一个栈实现正常的push、pop、top，另一个栈的栈顶每次都只放以一个栈中最小的元素 

class MinStack {

    private Stack<Integer> stack1;
    private Stack<Integer> stack2;

    public MinStack() {
        stack1 = new Stack<>();
        stack2 = new Stack<>();
    }

    public void push(int val) {
        stack1.push(val);
        if (stack2.isEmpty()) {
            stack2.push(val);
        } else {
            stack2.push(val < stack2.peek() ? val : stack2.peek());
        }
    }

    public void pop() {
        stack1.pop();
        stack2.pop();
    }

    public int top() {
        return stack1.peek();
    }

    public int getMin() {
        return stack2.peek();
    }
}
/**
 * Your MinStack object will be instantiated and called as such:
 * MinStack obj = new MinStack();
 * obj.push(val);
 * obj.pop();
 * int param_3 = obj.top();
 * int param_4 = obj.getMin();
 */