LeetCode 232. Implement Queue using Stacks–C++解法
LeetCode題解專欄:LeetCode題解
我做的所有的LeetCode的題目都放在這個專欄裏,大部分題目Java和Python的解法都有。
題目地址:Implement Queue using Stacks - LeetCode
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.
Example:
MyQueue queue = new MyQueue();
queue.push(1);
queue.push(2);
queue.peek(); // returns 1
queue.pop(); // returns 1
queue.empty(); // returns false
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, 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).
這道題目是經典的用2個棧來實現一個隊列,達到入隊和出隊的時間複雜度都是O(1)。
C++解法如下:
class MyQueue {
public:
stack<int> inbox, outbox;
/** Initialize your data structure here. */
MyQueue() {
;
}
/** Push element x to the back of queue. */
void push(int x) {
inbox.push(x);
}
/** Removes the element from in front of queue and returns that element. */
int pop() {
int temp = peek();
outbox.pop();
return temp;
}
/** Get the front element. */
int peek() {
cout << outbox.empty() << endl;
if (outbox.empty() == 1) {
while (inbox.empty() == 0) {
outbox.push(inbox.top());
inbox.pop();
}
}
return outbox.top();
}
/** Returns whether the queue is empty. */
bool empty() {
return inbox.empty() && outbox.empty();
}
};
/**
* Your MyQueue object will be instantiated and called as such:
* MyQueue* obj = new MyQueue();
* obj->push(x);
* int param_2 = obj->pop();
* int param_3 = obj->peek();
* bool param_4 = obj->empty();
*/