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Job Hunting Interview
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1

印度小哥斯坦福但是没有口音！和之前一位仁兄发过的面经一模一样！就是p e r s i s t a n t 堆 s t ac k
但是我实现下来感觉没有那么复杂 就类似用个linkedlist一样的structure就行了
需要设计constructor，peek，pop，size
然后还要test，run最后好不容易过了哈哈

2

想請教lz什麼是persistant 堆 stack？
用linkedlist 來實作stack嗎？還是什麼
感謝lz

3

你给个邮箱我直接全部发给你吧

4

http://www.1point3acres.com/bbs/thread-233854-1-1.html

补充内容 (2018-2-28 04:50):
参考这位写的帖子 描述的很详细了

5

好的 我來看一下：）

6

我和楼主感觉差不多呀 是不是那个仁兄想复杂了

7

lz最后结果怎么样呢

8

求问楼主电面怎么准备， 有没有靠谱面经哇。。。下周面试，没想到这家这么难

9

我写了个python的版本

import unittest


class PStack:
    def __init__(self, prev=None, value=None, size=0):
        self.prev = prev
        self.value = value
        self.size = size

    def push(self, x):
        return PStack(self, x, self.size + 1)

    def pop(self):
        return self.prev

    def peek(self):
        return self.value

    def getSize(self):
        return self.size


class PStackTest(unittest.TestCase):
    def testPstack(self):
        pstack = PStack()

        s1 = pstack.push(1)
        s2 = pstack.push(2)
        s3 = s2.push(3)
        tmp = s3.pop()

        self.assertEqual(0, pstack.getSize())
        self.assertEqual(1, s1.getSize())
        self.assertEqual(1, s1.peek())
        self.assertEqual(1, s2.getSize())
        self.assertEqual(2, s2.peek())
        self.assertEqual(3, s3.peek())
        self.assertEqual(2, s3.getSize())
        self.assertEqual(tmp, s2)


if __name__ == "__main__":
    unittest.main()
10

最新 电面

Given an input list of names, for each name, find the shortest substring that only appears in that name.

Example:

Input: ["cheapair", "cheapoair", "peloton", "pelican"]
Output:
"cheapair": "pa"  // every other 1-2 length substring overlaps with cheapoair
"cheapoair": "po" // "oa" would also be acceptable
"pelican": "ca"   // "li", "ic", or "an" would also be acceptable
"peloton": "t"    // this single letter doesn't occur in any other string