[Golang][Leetcode][Stack & Queue]刷題系列-225-Implement Stack using Queues

Contents
225. Implement Stack using Queues
Level : Easy
原題連結 : Click
題目 :
Implement a last-in-first-out (LIFO) stack using only two queues. The implemented stack should support all the functions of a normal stack (push, top, pop, and empty).
Implement the MyStack class:
- void push(int x) Pushes element x to the top of the stack.
- int pop() Removes the element on the top of the stack and returns it.
- int top() Returns the element on the top of the stack.
- boolean empty() Returns true if the stack is empty, false otherwise.
Notes:
- You must use only standard operations of a queue, which means that only push to back, peek/pop from front, size and is empty operations are valid.
- Depending on your language, the queue may not be supported natively. You may simulate a queue using a list or deque (double-ended queue) as long as you use only a queue’s standard operations.
Example :
Note
Input [“MyStack”, “push”, “push”, “top”, “pop”, “empty”]
[[], [1], [2], [], [], []]
Output
[null, null, null, 2, 2, false]
Explanation
MyStack myStack = new MyStack();
myStack.push(1);
myStack.push(2);
myStack.top(); // return 2
myStack.pop(); // return 2
myStack.empty(); // return False
解題思路 :
- 這題其實題目看似有些繁雜但是其實想法非常簡單
- 只要釐清Stacks是FILO(first in last out),Queues是FIFO(fisrt in first out)的差別
- 再來需要考慮我們需要多少queue去實現一個stack,其實呢一個就夠了
- 我們push操作用append去做處理不斷的往後加,top就是使用提取最後一個元素出來也就是queue[len(queue)-1],pop需要刪除stack內的元素,實現方法就是用golang的slice方法queue[:len(queue)-1],就可以成功完成,easy!!
以下是我的解法
- 若要探討每一個function的時間複雜度的話,需要先了解所謂golang的 append 背後的原理,其實呢,append其實是根據其放入的slice他的base array的capacity來決定是否需要新allocate一個array,所以它的時間複雜度best case 會在O(1),worst case 會在O(n)
- 所以push, pop, top也是best case O(1),worst case O(n)
Runtime: 0 ms, faster than 100.00% of Go online submissions for Implement Stack using Queues.
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