最小栈

题目

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

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

输出:
[null,null,null,null,-3,null,0,-2]

解释:
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.

题解

原理就是维护一个普通的栈和一个只能插入最小数字的栈, 这样通过 length 就能获取到最小的数字, 以达到线性时间检索到最小元素的目的.

/**
 * initialize your data structure here.
 */
var MinStack = function () {
  this.items = []
  this.minItems = []
}

/**
 * @param {number} x
 * @return {void}
 */
MinStack.prototype.push = function (x) {
  this.items.push(x)

  if (this.minItems.length === 0 || x <= this.getMin()) {
    this.minItems.push(x)
  }
}

/**
 * @return {void}
 */
MinStack.prototype.pop = function () {
  if (this.top() === this.getMin()) {
    this.minItems.pop()
  }

  this.items.pop()
}

/**
 * @return {number}
 */
MinStack.prototype.top = function () {
  return this.items[this.items.length - 1]
}

/**
 * @return {number}
 */
MinStack.prototype.getMin = function () {
  return this.minItems[this.minItems.length - 1]
}

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