[leetcode] 232. Implement Queue using Stacks 解题报告

题目链接:https://leetcode.com/problems/implement-queue-using-stacks/

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 toppeek/pop from topsize, 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).

思路:用两个栈分别叫做st1、st2,来模拟一个队列。对于以下操作的思想是:

1. pop -- 我们知道栈的特性是后入先出,队列的特性是先入先出。所以要出队列的值是最先入队列的,而这个值在st1的最底部,因此我们可以将栈st1的值全部取出,放入栈st2中,这样在栈st2顶部的值就是队列的首部。那么当我们要出队列一个值的时候如果st2不为空,则可知其栈顶元素就是我们队列的首个元素。因此把这个元素出栈即可。

2. peek -- 和pop一样,取栈顶元素,但不出栈。

3. push -- 知道了上两个操作,这个就很简单了,直接放进st1即可。

4. empty -- 如果两个栈都为空则队列为空。


代码如下:

class Queue {
public:
    // Push element x to the back of queue.
    void push(int x) {
        st1.push(x);
    }

    // Removes the element from in front of queue.
    void pop(void) {
        if(!st2.empty())
        {
            st2.pop();
            return;
        }
        while(!st1.empty())
        {
            st2.push(st1.top());
            st1.pop();
        }
        st2.pop();
    }

    // Get the front element.
    int peek(void) {
        if(!st2.empty())
            return st2.top();
        while(!st1.empty())
        {
            st2.push(st1.top());
            st1.pop();
        }
        return st2.top();
    }

    // Return whether the queue is empty.
    bool empty(void) {
        return st1.empty() && st2.empty();
    }
private:
    stack<int> st1;
    stack<int> st2;
};

你可能感兴趣的:(LeetCode,算法,Queue,stack)