155. Min Stack 155.最小栈

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 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

思路：还是有点不好想的。只用一个stack和一个min变量。push两次，pop两次。
push两次：push一个以前的min作为垫背。就是push也留个备胎吧，毕竟要pop两次
pop两次：min == stack.pop()出去了，所以min要更新为当前的值

🌰：

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

push max
push -2

push 0

push -2 
push -3

pop -3 min = -2 更新了, -2也pop出去了

top = 2
min = -2

class MinStack {
    int min = Integer.MAX_VALUE;
    Stack<Integer> stack = new Stack<Integer>();
    public void push(int x) {
        //push一个以前的min作为垫背
        // only push the old minimum value when the current 
        // minimum value changes after pushing the new value x
        if(x <= min){          
            stack.push(min);
            min=x;
        }
        stack.push(x);
    }

    public void pop() {
        // if pop operation could result in the changing of the current minimum value, 
        // pop twice and change the current minimum value to the last minimum value.
        if(stack.pop() == min) min=stack.pop();
    }

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

    public int getMin() {
        return min;
    }
}