232. Implement Queue using Stacks

Level : Easy

原題連結 : Click

題目 :

Implement a first in first out (FIFO) queue using only two stacks. The implemented queue should support all the functions of a normal queue (push, peek, pop, and empty).

Implement the MyQueue class:

• void push(int x) Pushes element x to the back of the queue.
• int pop() Removes the element from the front of the queue and returns it.
• int peek() Returns the element at the front of the queue.
• boolean empty() Returns true if the queue is empty, false otherwise.

Notes:

• You must use only standard operations of a stack, which means only push to top, peek/pop from top, size, and is empty operations are valid.
• Depending on your language, the stack may not be supported natively. You may simulate a stack using a list or deque (double-ended queue) as long as you use only a stack’s standard operations.

Example :

解題思路 :

• 這題需要用兩個stack來模擬出queue FIFO(First in first out)的情況，stack本身就是一種先進後出，單一出口的資料結構，若要模擬出queue可以有一個stack-in跟stack-out

• 一個負責接收element(push)，一個負責釋放element(pop/peek)

  1. push的實作非常簡單，就是直接將element append進stack-in array
  2. peek/pop的實作我用了一些邏輯，需要先將stack-in中的element搬到stack-out，這邊需要注意取出的方式是stack-in[len(stack-in)-1]，然後append到stack-out這樣的取出方式可以達到將stack-in排序方式翻轉的效果，接下來再從stack-out用同樣方式取出時就符合FIFO的特性

以下是我的解法

• 若要探討每一個function的時間複雜度的話，需要先了解所謂golang的 append 背後的原理，其實呢，append其實是根據其放入的slice他的base array的capacity來決定是否需要新allocate一個array，所以它的時間複雜度best case 會在O(1)，worst case 會在O(n)
• 所以push, pop, peek也是best case O(1)，worst case O(n)

Runtime: 0 ms, faster than 100.00% of Go online submissions for Implement Queue using Stacks.

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type MyQueue struct {
    stack_in []int
    stack_out []int
}


/** Initialize your data structure here. */
func Constructor() MyQueue {
    return MyQueue {
        stack_in : make([]int, 0),
        stack_out : make([]int, 0),
    }
}


/** Push element x to the back of queue. */
func (this *MyQueue) Push(x int)  {
    this.stack_in = append(this.stack_in, x)
    
}


/** Removes the element from in front of queue and returns that element. */
func (this *MyQueue) Pop() int {
    val := this.Peek()
    this.stack_out = this.stack_out[:len(this.stack_out)-1]
    
    return val
}


/** Get the front element. */
func (this *MyQueue) Peek() int {
    if len(this.stack_out) == 0 {
        for len(this.stack_in) > 0 {
            val := this.stack_in[len(this.stack_in)-1]
            this.stack_in = this.stack_in[:len(this.stack_in)-1]
            this.stack_out = append(this.stack_out,val)
        }
    }
    return this.stack_out[len(this.stack_out)-1]
}


/** Returns whether the queue is empty. */
func (this *MyQueue) Empty() bool {
    return len(this.stack_in) == 0 && len(this.stack_out) == 0
}


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

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