大纲
- 题目
- 地址
- 内容
- 解题
- 代码地址
题目
地址
https://leetcode.com/problems/implement-queue-using-stacks/description/
内容
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
, andis 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.
Example 1:
Input
[“MyQueue”, “push”, “push”, “peek”, “pop”, “empty”]
[[], [1], [2], [], [], []]
Output
[null, null, null, 1, 1, false]
Explanation
MyQueue myQueue = new MyQueue();
myQueue.push(1); // queue is: [1]
myQueue.push(2); // queue is: [1, 2] (leftmost is front of the queue)
myQueue.peek(); // return 1
myQueue.pop(); // return 1, queue is [2]
myQueue.empty(); // return false
Constraints:
1 <= x <= 9
- At most
100
calls will be made topush
,pop
,peek
, andempty
. - All the calls to
pop
andpeek
are valid.
Follow-up: Can you implement the queue such that each operation is amortized 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.
解题
这题和《闯关leetcode——225. Implement Stack using Queues》非常类似。它要求使用两个栈表达出队列的操作。栈的特点是先进后出,队列的特点是先进先出。我们将借助一个栈来做中转。
#include <stack>
using namespace std;
class MyQueue {
public:
MyQueue() {
}
void push(int x) {
s1.push(x);
}
int pop() {
int res = 0;
while (!s1.empty()) {
s2.push(s1.top());
s1.pop();
}
res = s2.top();
s2.pop();
while (!s2.empty()) {
s1.push(s2.top());
s2.pop();
}
return res;
}
int peek() {
int res = 0;
while (!s1.empty()) {
s2.push(s1.top());
s1.pop();
}
res = s2.top();
while (!s2.empty()) {
s1.push(s2.top());
s2.pop();
}
return res;
}
bool empty() {
return s1.empty();
}
private:
stack<int> s1, s2;
};
代码地址
https://github.com/f304646673/leetcode/blob/main/232-Implement-Queue-using-Stacks/cplusplus/src/solution.hpp