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implement-stack-using-queues(easy,但也有思考价值)
https://leetcode.com/problems/implement-stack-using-queues/
还有种方法,就是利用同一个队列,知道队列长度前提下,把内容从头到尾,再向尾部依次重推一下。
package com.company; import java.util.Deque; import java.util.LinkedList; import java.util.Queue; class Solution { Queue[] queues; int cur; Solution() { queues = new LinkedList[2]; queues[0] = new LinkedList<Integer>(); queues[1] = new LinkedList<Integer>(); cur = 0; } // Push element x onto stack. public void push(int x) { queues[cur].offer(x); } // Removes the element on top of the stack. public void pop() { change(true); } // Get the top element. public int top() { return change(false); } private int change(boolean pop) { int next = (cur + 1) % 2; int num = (int)queues[cur].poll(); while (!queues[cur].isEmpty()) { queues[next].offer(num); num = (int)queues[cur].poll(); } if (!pop) { queues[next].offer(num); } cur = next; return num; } // Return whether the stack is empty. public boolean empty() { return queues[cur].isEmpty(); } } public class Main { public static void main(String[] args) throws InterruptedException { System.out.println("Hello!"); Solution solution = new Solution(); // Your Codec object will be instantiated and called as such: solution.push(1); solution.push(2); solution.push(3); int ret = solution.top(); System.out.printf("ret:%d\n", ret); solution.pop(); ret = solution.top(); System.out.printf("ret:%d\n", ret); solution.pop(); ret = solution.top(); System.out.printf("ret:%d\n", ret); System.out.println(); } }
implement-stack-using-queues(easy,但也有思考价值)
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