Implement Queue by Two Stacks
As the title described, you should only use two stacks to implement a queue's actions.
The queue should supportpush(element)
,pop()
andtop()
where pop is pop the first(a.k.a front) element in the queue.
Both pop and top methods should return the value of first element.
Example
push(1)
pop() // return 1
push(2)
push(3)
top() // return 2
pop() // return 2
分析
一个栈进,一个栈出,每次pop或者top时候,若s2空,s1都倒入s2
public class MyQueue {
private Stack<Integer> stack1;
private Stack<Integer> stack2;
public MyQueue() {
// do intialization if necessary
stack1 = new Stack<Integer>();
stack2 = new Stack<Integer>();
}
/*
* @param element: An integer
* @return: nothing
*/
public void push(int element) {
// write your code here
stack1.push(element);
}
/*
* @return: An integer
*/
public int pop() {
// write your code here
if(!stack2.isEmpty()){
return stack2.pop();
}else{
while(!stack1.isEmpty()){
stack2.push(stack1.pop());
}
}
return stack2.pop();
}
/*
* @return: An integer
*/
public int top() {
// write your code here
if(!stack2.isEmpty()){
return stack2.peek();
}else{
while(!stack1.isEmpty()){
stack2.push(stack1.pop());
}
}
return stack2.peek();
}
}
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