interview fb2

2014.7.8fb

#include <iostream>
using namespace std;

struct TreeNode{
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int val)
    {
        left = NULL;
        right = NULL;
    }
};

/*

Input : binary tree

      A
     /     B   C
   / \     D   E   G
       \           F   H

Output :
A
BC
DEG
FH
*/
vector<vector<int> > transform(TreeNode *root)
{
    vector<vector<int> ans;
    if(root == NULL)
        return ans;

    vector<int> ans_piece;
    queue<TreeNode *> last_time;

    queue.push_back(root);
    TreeNode *temp;

    vector<int> num_each;
    num_each.push_back(1);
    num_each.push_back(0);
    int level = 0;
    while(queue.empty() == false)
    {
        num_each[level]--;

        temp = queue.get_first_one();
        queue.pop();
        if(temp)
            ans_piece.push_back(temp->val);
        if(temp->left)
        {
            queue.push_back(temp->left);
            num_each[level+1]++;
        }
        if(temp->right)
        {
            queue.push_back(temp->right);
            num_each[level+1]++;
        }

        if(num_each[level] == 0)
        {
            ans.push_back(ans_piece);
            level++;
            ans_piece.clear();
            num_each.push_back(0);
        }
    }
    return ans;
}
 

interview fb2,布布扣,bubuko.com

时间: 2024-07-31 14:31:33

interview fb2的相关文章

《Cracking the Coding Interview》——第16章:线程与锁——题目5

2014-04-27 20:16 题目:假设一个类Foo有三个公有的成员方法first().second().third().请用锁的方法来控制调用行为,使得他们的执行循序总是遵从first.second.third的顺序. 解法:你应该想到了用lock的方法类阻塞,不过这里面有个概念问题使得直接用ReentrantLock不能通过编译(对于一个锁对象,不同在A线程中锁定,又在B线程中解锁,不允许这样的归属关系),可以用Semaphore来达到相同的目的.请看下面的代码. 代码: 1 // 16

《Cracking the Coding Interview》——第16章:线程与锁——题目3

2014-04-27 19:26 题目:哲学家吃饭问题,死锁问题经典模型(专门用来黑哲学家的?). 解法:死锁四条件:1. 资源互斥.2. 请求保持.3. 非抢占.4. 循环等待.所以,某砖家拿起一只筷子后如果发现没有另一只了,就必须把手里这只筷子放下,这应该是通过破坏"请求保持"原则来防止死锁产生,请求资源失败时,连自己的资源也进一步释放,然后在下一轮里继续请求,直到成功执行. 代码: 1 // This is the class for chopsticks. 2 import j

《Cracking the Coding Interview》——第16章:线程与锁——题目2

2014-04-27 19:14 题目:如何测量上下文切换的时间? 解法:首先,上下文切换是什么,一搜就知道.对于这么一个极短的时间,要测量的话,可以通过放大N倍的方法.比如:有A和B两件事,并且经常一起发生,每件只需要花几纳秒.如果你把A事件连续做几百万次,而B时间只做了几次,这样就能排除B事件对于测量的影响.如果总时间S = mA + nB.当m >> n 时,A≈S / m.下面的测量方法类似于打乒乓球,在主线程和副线程间互相传递一个令牌,这个令牌可以是变量.管道之类的用于通信的工具.与

《Cracking the Coding Interview》——第16章:线程与锁——题目1

2014-04-27 19:09 题目:线程和进程有什么区别? 解法:理论题,操作系统教材上应该有很详细的解释.我回忆了一下,写了如下几点. 代码: 1 // 16.1 What is the difference between process and thread? 2 Answer: 3 Process: 4 1. Basic element of resource allocation in the operating system. 5 2. Possesses independent

Popular HashMap and ConcurrentHashMap Interview Questions

http://howtodoinjava.com/core-java/collections/popular-hashmap-and-concurrenthashmap-interview-questions/ Popular HashMap and ConcurrentHashMap Interview Questions June 14, 2013 by Lokesh Gupta In my previous post related to “How HashMap works in jav

Today&#39;s interview of C#

I think they are advanced topics. C# 1. when will you use list, when will you use hashtable. 2. when will you use Idispose. 3. Async and Await public async Task<string> DoAsync5()        {           Task<string> mytask=Task.Run<string>((

Java Interview Reference Guide--reference

Part 1 http://techmytalk.com/2014/01/24/java-interview-reference-guide-part-1/ Posted on January 24, 2014 by Nitin Kumar JAVA Object Oriented Concepts Java in based on Object Oriented concepts, which permits higher level of abstraction to solve any p

《Cracking the Coding Interview》——第18章:难题——题目10

2014-04-29 04:22 题目:给定一堆长度都相等的单词,和起点.终点两个单词,请从这堆单词中寻找一条变换路径,把起点词变成终点词,要求每次变换只能改一个字母. 解法:Leetcode中有Word Ladder,这题基本思路一致. 代码: 1 // 18.10 Given a list of words, all of same length. Given a source and a destionation words, you have to check if there exis

《Cracking the Coding Interview》——第18章:难题——题目9

2014-04-29 04:18 题目:有一连串的数被读入,设计一个数据结构,能随时返回当前所有数的中位数. 解法:用一个大顶堆,一个小顶堆将数分成数量最接近的两份,就能轻松得到中位数了. 代码: 1 // 18.9 A stream of integers are passed to you, you have to tell me the median as they keep coming in. 2 #include <climits> 3 #include <iostream&