amazon 面经# JobHunting - 待字闺中
i*t
1 楼
first phone interview:
why amazon
java hibernate
DOS, DDOS (security)
写SQL 一堆
写程序: given a unsorted integer array, find two number which sum is equal
to the given number. (integer may be negative)
second phone interview:
describe hardest problem you met
java里的 final, finally, finalize
java的static
如何判断给定的地址是big-endian还是little-endian?
写程序: count the number of 0 bit in a given string
写程序: x power of y
刚面完第二次phone interview, 太紧张, 答的不好
bless me~~~
why amazon
java hibernate
DOS, DDOS (security)
写SQL 一堆
写程序: given a unsorted integer array, find two number which sum is equal
to the given number. (integer may be negative)
second phone interview:
describe hardest problem you met
java里的 final, finally, finalize
java的static
如何判断给定的地址是big-endian还是little-endian?
写程序: count the number of 0 bit in a given string
写程序: x power of y
刚面完第二次phone interview, 太紧张, 答的不好
bless me~~~