[Offer] to prove safety 05 - queue using two stacks

Reconstruction of a binary tree

Time limit : 1秒
space constraints : 32768K
this question knowledge : 队列 栈
Title Description :

用两个栈来实现一个队列，完成队列的Push和Pop操作。 队列中的元素为int类型。

import java.util.Stack;

public class Solution {
    Stack<Integer> stack1 = new Stack<Integer>();
    Stack<Integer> stack2 = new Stack<Integer>();
    
    public void push(int node) {
        
    }
    
    public int pop() {
    
    }
}

Analysis of ideas:

Operation of sequentially undergoes the following changes:

1. Two stacks are empty

2. -> stack1 not empty empty stack2

3. -> stack1 empty stack2 not empty

4. -> two stacks are not empty

Two stacks are empty

Two stacks are empty	stack1=[]	stack2=[]	Steps
Drawing {1, 2}	stack1=[2, 1]	stack2=[]	2 * stack1.push
Pop	stack1=[]	stack2=[]	return null

stack1 not empty stack2 empty

1 2 not empty empty	stack1=[2, 1]	stack2=[]	Steps
Drawing {3, 4}	stack1=[4, 3, 2, 1]	stack2=[]	2 * stack1.push
Pop	stack1=[]	stack2=[2, 3, 4]	4 * stack1.pop 4 * stack2.push return stack2.pop

stack1 empty stack2 not empty

1 2 not empty Empty	stack1=[]	stack2=[2, 3, 4]	Steps
Drawing {5, 6}	stack1=[6, 5]	stack2=[2, 3, 4]	2 * stack1.push
Pop	stack1=[]	stack2=[3, 4]	return stack2.pop

Two stack is not empty

Two stack is not empty	stack1=[6, 5]	stack2=[2, 3, 4]	Steps
Drawing {7, 8}	stack1=[8, 7, 6, 5]	stack2=[2, 3, 4]	2 * stack1.push
Pop	stack1=[6, 5]	stack2=[3, 4]	return stack2.pop

summary:

1. Stack situation: always stack1.push
2. The stack case:
   • stack2 empty: all the elements of the stack stack1 to stack2, then perform stack2.pop
   • stack2 not empty: stack2.pop

answer:

import java.util.Stack;

public class Solution {
    Stack<Integer> stack1 = new Stack<Integer>();
    Stack<Integer> stack2 = new Stack<Integer>();
    
    public void push(int node) {
        stack1.push(node);
    }
    
    public int pop() {
        // 若 stack2 为空：将所有 stack1 元素入栈到 stack2
        if(stack2.empty()){
            while(!stack1.empty()){
                stack2.push(stack1.pop());
            }
        }
        return stack2.pop();
    }
}