（LeetCode）用兩個棧實現一個隊列
阿新 • • 發佈：2017-08-18
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LeetCode上面的一道題目。原文例如以下：
Implement the following operations of a queue using stacks.
- push(x) -- Push element x to the back of queue.
- pop() -- Removes the element from in front of queue.
- peek() -- Get the front element.
- empty() -- Return whether the queue is empty.
- You must use onlystandard operations of a stack -- which means only
push to top,peek/pop from top,size, andis emptyoperations are valid. - Depending on your language, stack may not be supported natively. You may simulate a stack by using a list or deque (double-ended queue), as long as you use only standard operations of a stack.
- You may assume that all operations are valid (for example, no pop or peek operations will be called on an empty queue).
代碼例如以下：
class MyQueue {
// Push element x to the back of queue.
Stack<Integer> stack1 = new Stack<Integer>();
Stack<Integer> stack2 = new Stack<Integer>();
public void push(int x) {
stack1.push(x);
}
// Removes the element from in front of queue.
public void pop() {
if(stack2.size()==0)
{
int m = stack1.size();
for(int i=0;i<m;i++)
{
stack2.push(stack1.pop());
}
}
stack2.pop();
}
// Get the front element.
public int peek() {
if(stack2.size()==0)
{
int m = stack1.size();
for(int i=0;i<m;i++)
{
stack2.push(stack1.pop());
}
}
return stack2.peek();
}
// Return whether the queue is empty.
public boolean empty() {
return stack1.size()==0&&stack2.size()==0;
}
}（LeetCode）用兩個棧實現一個隊列