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.

Solution

在用一个单独的stack记录min值的方法上做个优化，只用一个stack才存储：
每次更新min值的时候，把当前值cur push进来之前的min值也push进stack，这样每次pop出cur都能够马上知道剩下的数中min值是多少。
pop的时候，如果peek等于当前min值，pop出来之后的peek就是剩下数的min值

class MinStack {

    Stack<Integer> stack;
    int min;
    /** initialize your data structure here. */
    public MinStack() {
        stack = new Stack<Integer>();
        min = Integer.MAX_VALUE;
    }

    public void push(int x) {
        if (stack.isEmpty() || min >= x) {
            stack.push(min);
            min = x;
        }

        stack.push(x);
    }

    public void pop() {
        int cur = stack.pop();
        if (cur == min) {
            min = stack.pop();
        }
    }

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

    public int getMin() {
        return 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();
 */