Comments for Zhu Xinquan's Blog

Comment on Notes of Introduction to Algorithm(B-Tree) by Mike McNally

Mike McNally — Sat, 21 Jul 2007 06:42:57 +0000

Hi. I’m no expert, but as I understand it a “2-3″ tree (each node has either one or two entries and two or three (or none) children) are also B-Trees.

Comment on First home in Shanghai by Gajendra

Gajendra — Thu, 19 Jul 2007 21:49:15 +0000

nice room & really you have placed all the thing on the suitable palce.

Comment on Job Opportunity at Morgan Stanley by Leon

Leon — Fri, 06 Jul 2007 15:06:27 +0000

How’s the salary go? I know it’s open for the capability, but how much is it?

Comment on Introduction to Algorithms by xinquan

xinquan — Thu, 05 Jul 2007 21:47:11 +0000

Oronno,
Sorry, I　don’t know any other link yet. If you find in the future, please tell me as well.

Comment on Introduction to Algorithms by Oronno

Oronno — Tue, 26 Jun 2007 12:15:19 +0000

Could you please give me another link of CLR’s solution.
Cause the link that u gave for solution is broken..
Thanks..

Comment on Autoconf Macro Reference by budowa domów

budowa domów — Sun, 24 Jun 2007 18:53:16 +0000

Good article and site. Congratulations

Comment on Happy Birthday to Myself again by jjgirl

jjgirl — Tue, 05 Jun 2007 11:49:00 +0000

Hei, I don’t know what’s the matter with you.
but I believe you can deal with everything.
Man, you are gonna change, cause all of your friends are watching you for they want you a better life.
For fire, for love, for dead, for live, then it happens, the dream will comes true.
blessing~~

Comment on Happy Birthday to Myself again by Hu

Hu — Fri, 18 May 2007 00:47:28 +0000

HAPPY BIRTHDAY

Comment on Notes of Introduction to Algorithm(Exercise 2.3.7) by 金海

金海 — Sat, 12 May 2007 23:43:42 +0000

Your code works.
I personally found that the following technique works as well.
First we sort the array using merge sort, which is O(n lg n) in time;
Then, we apply the following procedure
i = 0;
j = n-1; // size of array is n
while (j>i)
if ( a[i] + a[j] == x )
output message and return;
else if ( a[i] + a[j] > x )
–j; // decrease j
else
++i; // increase i;
Each time we reduce the size of the array by 1. So I claim the above procedure is O(n) in time.
Third, totally, it is O( n lgn ) + O (n) = O( n lgn).
Complete code is pasted below and thanks for posting your work.
#include
#include “hjns.h” // for DIM
using namespace std;
void merge(int* a, int p, int q, int r);
void merge_sort(int*a, int p, int r);
void sum(int *a, int n, int x);
int main(int argc, char** argv)
{
int a[] = {1, 4, 9, -10, 8, 5};
int x = -1;
sum(a, DIM(a), x);
return 0;
}
void merge(int* a, int p, int q, int r)
{
int i=p;
int j=q+1;
int k=0;
int* b = new int[r-p+1];
while(ii)
{
if(a[i]+a[j] == x)
{
cout x)
–j;
else
++i;
}
cout<“no two elements add to “< }

Comment on Notes of Introduction to Algorithm(Order Statistics) by xiaozhu

xiaozhu — Sun, 25 Mar 2007 22:41:16 +0000

correct