Golang solution shifting elements from one stack to another on-demand


  • 0

    The idea is pretty much borrowed from Cracking The Code Interview.

    type MyQueue struct {
    	Stack *[]int
    	Queue *[]int
    }
    
    /** Initialize your data structure here. */
    func Constructor() MyQueue {
    	tmp1, tmp2 := []int{}, []int{}
    	return MyQueue{Stack: &tmp1, Queue: &tmp2}
    }
    
    /** Push element x to the back of queue. */
    func (this *MyQueue) Push(x int) {
    	*this.Stack = append(*this.Stack, x)
    }
    
    /** Removes the element from in front of queue and returns that element. */
    func (this *MyQueue) Pop() int {
    	if len(*this.Queue) == 0 {
    		this.fromStackToQueue(this.Stack, this.Queue)
    	}
    
    	popped := (*this.Queue)[len(*this.Queue)-1]
    	*this.Queue = (*this.Queue)[:len(*this.Queue)-1]
    	return popped
    }
    
    /** Get the front element. */
    func (this *MyQueue) Peek() int {
    	if len(*this.Queue) == 0 {
    		this.fromStackToQueue(this.Stack, this.Queue)
    	}
    
    	return (*this.Queue)[len(*this.Queue)-1]
    }
    
    /** Returns whether the queue is empty. */
    func (this *MyQueue) Empty() bool {
    	return len(*this.Stack)+len(*this.Queue) == 0
    }
    
    func (this *MyQueue) fromStackToQueue(s, q *[]int) {
    	for len(*s) > 0 {
    		popped := (*s)[len(*s)-1]
    		*s = (*s)[:len(*s)-1]
    
    		*q = append(*q, popped)
    	}
    }
    

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