【第三轮】6.30-7.6 CareerCup 3.5

【解题思路】
enqueue: push to the pushStack
dequeue: if popStack is empty, (1) if pushStack is empty:EXCEPTION, (2) pop pushStack and push to popStack. pop one element from popStack.
【时间复杂度】
enqueue: O(1)
dequeue: O(N)
【空间复杂度】
O(N)
【gist link】
https://gist.github.com/iwilbert/6c6744b3ad423ad6a436

【解题思路】use 2 stacks , one for push, one for pop
【时间复杂度】o(n)
【空间复杂度】o(n)
【gist link】https://gist.github.com/hilda8519/50757a448bbfd7463a64

【解题思路】
using two stacks stackIn and stackOut to push/pop elements
push: stackIn.push
pop: if stackOut is empty, move all elements of stackIn into stackOut
size: stackIn.size + stackOut.size
back: stackIn.peek, if stackIn is empty, move all elements of stackOut into stackIn
front: stack.Out.peek, if stackOut is empty, move all elements of stackIn into stackOut
【时间复杂度】O(1) or O(n) (if the stack is empty)
【空间复杂度】O(N)
https://gist.github.com/weazord/f231350faa96658a1016

https://gist.github.com/weazord/f231350faa96658a1016

打了这么多字突然回复框内容全消失了, 不打了...

【解题思路】data_stack 和 buffer, pop和push来改变顺序，模拟FIFO
【时间复杂度】O(1) for enque; O(n) for deque()
【gist link】https://gist.github.com/xun-gong/5c1fe614008cdb762a8b

【解题思路】Create 2 stacks: one for add() and one for remove().
【时间复杂度】O(1) for add() and O(N) for remove().
【空间复杂度】O(N)
【gist link】https://gist.github.com/jason51122/ef61390dee4871050cda

【解题思路】
One stack for enqueue, one stack for dequeue
move elements to the other stack only when needed

【时间复杂度】
O(1) for best cases, O(n) for worst cases

【空间复杂度】
O(n)


【gist link】
https://gist.github.com/happyWinner/fc453f95338657706d05

【解题思路】
One stack for enqueue
The other stack for dequeue
Move all elements from one to the other to reverse the sequence if necessary
【时间复杂度】
O(N) in worst case

【空间复杂度】
O(n)

【gist link】
https://gist.github.com/Noahsark/62eb271bb42ff4d05c96