Implement Queue using Stacks - LeetCode
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.**Follow-up:** Can you implement the queue such that each operation is **[amortized](https://en.wikipedia.org/wiki/Amortized_analysis)** O(1) time complexity? In other words, performing n operations will take overall O(n) time even if one of those operations may take longer.
- code
class MyQueue:
def __init__(self):
self.i = []
self.o = []
def push(self, x):
self.i.append(x)
def pop(self):
if not self.o:
while self.i:
self.o.append(self.i.pop())
return self.o.pop()
def peek(self):
if not self.o:
while self.i:
self.o.append(self.i.pop())
return self.o[-1]
def empty(self):
return not self.i and not self.o
- code push O(1), pop amortized O(1)
class MyQueue:
def __init__(self):
self.s1 = []
self.s2 = []
def push(self, x):
self.s1.append(x)
def pop(self):
self.peek()
return self.s2.pop()
def peek(self):
if not self.s2:
while self.s1:
self.s2.append(self.s1.pop())
return self.s2[-1]
def empty(self):
return not self.s1 and not self.s2