因为之前发过贴，背景就不赘述了，具体说说怎么准备还有面经吧。
骑驴找马，一月底开始刷leetcode，到三月中第一个面试，刷了一遍半吧，明显觉得写
第二遍的时候思路清晰多了，code也比第一遍的简洁。其他的就是每家面试前争对性的
看面经，能看多少是多少，四家只有L面经重复率很高，g家最不能预料题型。后面准备
design的时候都是乱看，一些fb tech talk的视频还有之前有人贴过的fb design的总
结，
但我基础不好，临时抱佛脚感觉也没什么用。面经我就只贴面完有及时记下来的，反正
也给过很多朋友了，就贴上来吧。
已经签了fb，准备八月初start，有同一期的pm我，哈。
脸书:
1. Print all paths of a binary tree
I gave a recursive solution.
Then he wanted to give an iterative way.
2a. Fibonacci (iterative)
2b. Buckets of anagrams
[“cart”,”tarc”, “cat”, “act”, “ract”] -> [[“cart”, “tarc”, “
ract”], [“cat”, “act”]]
onsite design是tiny url, 估计interviewer也知道我没什么经验，问了个最简单的也
没答好。T-T
coding都比较easy。
领英:
1. Return if two strings are isomorphic. (character 1-1 match)
“zoo” -> “fee” ( z->f, o->e) true
“zoo” -> “dui” ( z->d, o->u, o-> ) false
“dui” -> “zoo” (d->z, u->o, i-> ) false
Use two hashmaps
*****************************************************************
2. K nearest points (solution see below) Time: O(nlgk)
*****************************************************************
1. Search in rotated sorted array
*****************************************************************
2. public interface Intervals {
/**
* Adds an interval [from, to] into internal structure.
*/
void addInterval(int from, int to);
/**
* Returns a total length covered by intervals.
* If several intervals intersect, intersection should be counted only
once.
* Example:
*
* addInterval(3, 6)
* addInterval(8, 9)
* addInterval(1, 5)
*
* getTotalCoveredLength() -> 6
* i.e. [1,5] and [3,6] intersect and give a total covered interval [1,6]
* [1,6] and [8,9] don't intersect so total covered length is a sum for
both intervals, that is 6.
*
* 0 1 2 3 4 5 6 7 8 9 10
*/
int getTotalCoveredLength();
}
亚麻:
1a. Given 2 sorted, singly-linked lists, write a function that will merge
them into a new sorted, singly-linked list
Ex.
1->2->4->8->16->32
2->4->6
1->2->2->4->4->6->8->16->32
*****************************************************************
1b. merge n sorted lists
// 1 -> 3,
// 2 -> 5
// 4
newhead: 1 -> 2 -> 3 -> 4 -> 5
*****************************************************************
1c. Given a Binary tree, print path from root to all nodes that are
divisible by 5
Input:
6
/
5 7
/
4 15
/ |
3 10 2 8
Output:
6 5
6 7 4 10
6 7 15
*****************************************************************
2. Given an array A (the array can be treated as a big number) and a number
n, find the biggest number that you can reach to via n swaps. A swap can
only happen in adjacent items. For example, given [1 3 4 2 5 7 9] and n = 1,
the biggest number is [3 1 4 2 5 7 9]
n=1, 3 1 4 2 5 7 9
n=2, 1 3 4 -> 1 4 3 -> 4 1 3
狗家:
1. Reorder List (leetcode)
1->2->3->4->5 => 1->5->2->4->3
*****************************************************************
2. Abbreviation: apple can be abbreviated to 5, a4, 4e, a3e, …
Given a target string (internationalization), and a set of strings,
return the minimal length of abbreviation of this target string so that it
won’t conflict with abbrs of the strings in the set.
“apple”, [“blade”] -> a4 (5 is conflicted with “blade”)
“apple”, [“plain”, “amber”, “blade”] -> ???
Problem changed to:
If given a string and an abbreviation, return if the string matches abbr.
“internationalization”, “i5a11o1” -> true
*****************************************************************
Onsite:
1a. Write a function to get a positive integer n as input and return 0 or 1.
The probability of returning 1 should be 1/(2^n)
1b. Given an array, return the median. (talk about expected time complexity)
2a. Code review - a class which takes a string, split by separators and
return the array of tokens (point out coding problems and indicate how you
will implement it)
2b. Longest consecutive sequence (leetcode) (how do you handle duplicates)
2c. design: how to store files given the file paths and contents. (tree?)
3a. Given an array and a number x, find out how many pairs satisfy (a[i], a[
j]) st. a[i]+a[j] < x
3b. follow up: if we want to find 3 items that adds up to a number < x
3c follow up: if we want to find k items. Time complexity: O(n^(k-1)*lgn)
4. Give a map which has some obstacles in it. Given a starting point S and
ending point E, find the shortest path from S to E. Note that you can go to
any(4) direction from S, but during the process, you can only go straight
from the previous direction, unless you hit an obstacle.
i.e. if you are at (1, 1) and the next (1, 2) is blocked, you can only go to
(2, 1) or (0, 1)
5a. Java “final” keyword
5b. 3-way partition: given an array and number x, reorder the array so that
first part will be < x, middle part is = x, and final part is > x.
5c. Design: given an array of integers and a range (i, j), we want to return
the min item in the range (balanced binary search tree)
5d. System design: given a machine, how to generate id so that they will not
duplicate; if we have multiple machines, what to do
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