题目
请用栈实现一个队列,支持如下四种操作:
push(x) – 将元素x插到队尾;
pop() – 将队首的元素弹出,并返回该元素;
peek() – 返回队首元素;
empty() – 返回队列是否为空;
注意:
你只能使用栈的标准操作:push to top,peek/pop from top, size 和 is empty;
如果你选择的编程语言没有栈的标准库,你可以使用list或者deque等模拟栈的操作;
输入数据保证合法,例如,在队列为空时,不会进行pop或者peek等操作;
样例
MyQueue queue = new MyQueue();
queue.push(1);
queue.push(2);
queue.peek(); // returns 1
queue.pop(); // returns 1
queue.empty(); // returns false
双队列模拟
很经典的题目了
设置两个栈in, out
push的时候把所有元素从out转移到in,然后再push
pop的时候把所有元素从in转移到out,然后再pop
class MyQueue {public:stack<int> in, out;/** Initialize your data structure here. */MyQueue() {}/** Push element x to the back of queue. */void push(int x) {while (!out.empty()) {in.push(out.top());out.pop();}in.push(x);}/** Removes the element from in front of queue and returns that element. */int pop() {while (!in.empty()) {out.push(in.top());in.pop();}int ans = out.top();out.pop();return ans;}/** Get the front element. */int peek() {while (!in.empty()) {out.push(in.top());in.pop();}return out.top();}/** Returns whether the queue is empty. */bool empty() {return in.empty() && out.empty();}};/*** Your MyQueue object will be instantiated and called as such:* MyQueue obj = MyQueue();* obj.push(x);* int param_2 = obj.pop();* int param_3 = obj.peek();* bool param_4 = obj.empty();*/
