包括一些书籍，文章等参考资料，内容比较杂
文件比较多，大家需要的话
留下10m以上的邮箱！
推荐大家java代码网址，经典！
http://www.javaref.cn
___
雅虎免费邮箱-3.5G容量，20M附件
http://cn.mail.yahoo.com/
--~--~-~--~~~---~--~~
You received this message
包括一些书籍，文章等参考资料，内容比较杂
文件比较多，大家需要的话
留下10m以上的邮箱！
推荐大家java代码网址，经典！
http://www.javaref.cn
___
Mp3疯狂搜-新歌热歌高速下
http://music.yahoo.com.cn/?source=mail_mailbox_footer
girsh and kamlesh: that may work if the (n-1) numbers are integers 1 to
n. What if they can be anything. (from your S(n) = n*(n+1) / 2 ) )
adak : distribution sort - this is effectively the same as the bitmap
thing which Lego Haryanto mentioned in the beginning. (we're only using
a byte or
yes, it's likely we think it's sorted since Joe uses the word
_sequence_ :
In a sequence o,f from 1 to n (1,2,3,...n) numbers, we need to find
the
missing numbers. There will not be duplicates.
anyway its a better problem if unsorted...
--~--~-~--~~~---~--~~
kumar wrote:
kerry wrote:
need O(n) time
the array is set[n],
for(i = 1, i =n ; i++)
if((set[i] 1) (set[i + 1] 1))
then miss the number i+1;
ur logic is not satisfied for even the array is in sorted order also.
For sorted array
we need ( set [ i ] 1) ( set [ i +
