题目描述

设计一个支持 push，pop，top 操作，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() -- 获取栈顶元素。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.

代码实现

/** * 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() */