LeetCode刷题实战232：用栈实现队列
共 3350字，需浏览 7分钟
· 2021-04-08
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).
示例
输入：
["MyQueue", "push", "push", "peek", "pop", "empty"]
[[], [1], [2], [], [], []]
输出：
[null, null, null, 1, 1, false]
解释：
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
解题
思路： 由于队列是先进先出的，而栈是先进后出的，所以要用2个栈来实现队列的入队出队功能，队列的入队功能与栈的一样，出队时，先将第一个栈中的元素全部弹出，并倒入到第二个栈中，将第二个栈中栈顶元素弹出，并将stack2中剩下的元素倒回到stack1中，即实现一次出队。
class MyQueue {
public:
/** Initialize your data structure here. */
stack<int> s1, s2;
MyQueue() {
}
/** Push element x to the back of queue. */
void push(int x) {
while(!s2.empty()) {
s1.push(s2.top());
s2.pop();
}
s2.push(x);
while(!s1.empty()) {
s2.push(s1.top());
s1.pop();
}
}
/** Removes the element from in front of queue and returns that element. */
int pop() {
int a = s2.top();
s2.pop();
return a;
}
/** Get the front element. */
int peek() {
return s2.top();
}
/** Returns whether the queue is empty. */
bool empty() {
return s2.empty();
}
};