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.
Notes:
- You must use only standard operations of a stack -- which means only
push to top
,peek/pop from top
,size
, andis empty
operations 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).
这道题让我们用栈来实现队列,之前我们做过一道相反的题目Implement Stack using Queues 用队列来实现栈,是用队列来实现栈。这道题颠倒了个顺序,起始并没有太大的区别,栈和队列的核心不同点就是栈是先进后出,而队列是先进先出,那么我们要用栈的先进后出的特性来模拟出队列的先进先出。那么怎么做呢,其实很简单,只要我们在插入元素的时候每次都都从前面插入即可,比如如果一个队列是1,2,3,4,那么我们在栈中保存为4,3,2,1,那么返回栈顶元素1,也就是队列的首元素,则问题迎刃而解。所以此题的难度是push函数,我们需要一个辅助栈tmp,把s的元素也逆着顺序存入tmp中,此时加入新元素x,再把tmp中的元素存回来,这样就是我们要的顺序了,其他三个操作也就直接调用栈的操作即可,参见代码如下:
class Queue { public: // Push element x to the back of queue. void push(int x) { stack<int> tmp; while (!s.empty()) { tmp.push(s.top()); s.pop(); } s.push(x); while (!tmp.empty()) { s.push(tmp.top()); tmp.pop(); } } // Removes the element from in front of queue. void pop(void) { s.pop(); } // Get the front element. int peek(void) { return s.top(); } // Return whether the queue is empty. bool empty(void) { return s.empty(); } private: stack<int> s; };
LeetCode All in One 题目讲解汇总(持续更新中...)
时间: 2024-10-29 03:43:21