问题形貌

设想一个支撑 push，pop，top 操纵，并能在常数时间内检索到最小元素的栈。

push(x) -- 将元素 x 推入栈中。
pop() -- 删除栈顶的元素。
top() -- 猎取栈顶元素。
getMin() -- 检索栈中的最小元素。

示例:

MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin();   --> 返回 -3.
minStack.pop();
minStack.top();      --> 返回 0.
minStack.getMin();   --> 返回 -2.

代码完成

/**
 * initialize your data structure here.
 */
var MinStack = function() {
  this.s1 = [];
  this.s2 = [];
};

/** 
 * @param {number} x
 * @return {void}
 */
MinStack.prototype.push = function(x) {
  let s2 = this.s2,
      s2Len = s2.length,
      s1 = this.s1;
  
  
  let curMin = s2[s2Len-1];
  if(curMin < x)
    s2.push(curMin);
  else
    s2.push(x);
  s1.push(x);
};

/**
 * @return {void}
 */
MinStack.prototype.pop = function() {
  let s1 = this.s1,
      s1Len = s1.length,
      s2 = this.s2,
      s2Len = s2.length;
  
  if(s1Len === 0)
    return undefined;
  
  s2.pop();
  return s1.pop();
};

/**
 * @return {number}
 */
MinStack.prototype.top = function() {
   let s1 = this.s1,
      s1Len = s1.length;
  if(s1.length === 0)
    return undefined;
  return s1[s1Len-1];
};

/**
 * @return {number}
 */
MinStack.prototype.getMin = function() {
  let s2 = this.s2,
      s2Len = s2.length;
  if(s2Len === 0)
    return undefined;
  
  return s2[s2Len-1];
};

/** 
 * Your MinStack object will be instantiated and called as such:
 * var obj = Object.create(MinStack).createNew()
 * obj.push(x)
 * obj.pop()
 * var param_3 = obj.top()
 * var param_4 = obj.getMin()
 */