225. 用队列实现栈

用队列实现栈

• 难度: easy
• 原始链接: https://leetcode.cn/problems/implement-stack-using-queues
• 标签: 队列, 堆栈

解法一: 双队列

go

type MyStack struct {
  queue1 *[]int

  queue2 *[]int
}

func Constructor() MyStack {
  return MyStack{
    queue1: &[]int{},
    queue2: &[]int{},
  }
}

func (this *MyStack) Push(x int) {
  queue1 := this.queue1
  queue2 := this.queue2
  *queue2 = append(*queue2, x)
  if len(*queue1) > 0 {
    *queue2 = append(*queue2, *queue1...)
    *queue1 = []int{}
  }
  this.queue1, this.queue2 = this.queue2, this.queue1
}

func (this *MyStack) Pop() int {
  queue1 := this.queue1
  ans := (*queue1)[0]
  (*queue1) = (*queue1)[1:]
  return ans
}

func (this *MyStack) Top() int {
  queue1 := this.queue1
  return (*queue1)[0]
}

func (this *MyStack) Empty() bool {
  queue1 := this.queue1
  return len(*queue1) == 0
}

/**
 * Your MyStack object will be instantiated and called as such:
 * obj := Constructor();
 * obj.Push(x);
 * param_2 := obj.Pop();
 * param_3 := obj.Top();
 * param_4 := obj.Empty();
 */

解法二: 一个队列

go

type MyStack struct {
  queue *[]int
}

func Constructor() MyStack {
  return MyStack{
    queue: &[]int{},
  }
}

func (this *MyStack) Push(x int) {
  queue := this.queue
  n := len(*queue)
  *queue = append(*queue, x)
  for i := 0; i < n; i++ {
    *queue = append(*queue, (*queue)[i])
  }
  *queue = (*queue)[n:]
}

func (this *MyStack) Pop() int {
  queue := this.queue
  ans := (*queue)[0]
  (*queue) = (*queue)[1:]
  return ans
}

func (this *MyStack) Top() int {
  queue := this.queue
  return (*queue)[0]
}

func (this *MyStack) Empty() bool {
  queue := this.queue
  return len(*queue) == 0
}

/**
 * Your MyStack object will be instantiated and called as such:
 * obj := Constructor();
 * obj.Push(x);
 * param_2 := obj.Pop();
 * param_3 := obj.Top();
 * param_4 := obj.Empty();
 */