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FinalFantasy — Wed, 22 Feb 2006 06:52:00 GMT

@�q�少无知
不知道你惌��什么样的东西，误��具体一些…�?img src ="http://www.aygfsteel.com/FinalFantasy/aggbug/31968.html" width = "1" height = "1" />

FinalFantasy 2006-02-22 14:52 发表评论

�q�少无知 — Sun, 19 Feb 2006 13:02:00 GMT

�l�我发点好的关于配置的东西嘛�?
wjz_86@126.com

�q�少无知 2006-02-19 21:02 发表评论

饉K�� — Thu, 05 Jan 2006 08:43:00 GMT

感觉不是那么回事
你的解释有点问题�Q?img src ="http://www.aygfsteel.com/FinalFantasy/aggbug/26735.html" width = "1" height = "1" />

饉K�� 2006-01-05 16:43 发表评论

re: Java回收机制分析

allenjiang — Tue, 27 Dec 2005 07:18:00 GMT

垃圾攉��是个老问题，也是一个研�I�的比较充分的问题，早在java、C�Q�＋语言出现前，�q�方面就已经有很多研�I�工作了�Q�记得有一本书��叫垃圾攉��Q�garbage collection�Q�，讲的比较详细
btw�Q�楼主讲的不错，谢谢

allenjiang 2005-12-27 15:18 发表评论

dotNet的程序员 — Thu, 22 Dec 2005 06:53:00 GMT

用栈改进一下�?
public class BusStops
{

public static void Main()
{
BusStops total = new BusStops();
//System.out.println(total.countStops({"...B.",".....","..X.B",".....","B...."},3));
System.Console.WriteLine(total.countStops(new string[]{"...B.",".....","..X.B",".....","B...."},3));
System.Console.ReadLine();
}

public int countStops(string[] cityMap, int walkingDistance)
{
Position xPosition=null;
System.Collections.Stack stack=new System.Collections.Stack();
for(int i=0;i {
for(int j=0;j {
if(cityMap[i][j]=='X')
{
xPosition=new Position();
xPosition.x = i;
xPosition.y = j;
}
else if(cityMap[i][j]=='B')
{
Position busStop=new Position();
busStop.x=i;
busStop.y=j;
stack.Push(busStop);
}
}
}
if (xPosition==null) throw new Exception("未发现X�?quot;);
if (stack.Count==0) return 0;
int sum= 0;
while (stack.Count>0)
{
Position busStop=(Position)stack.Pop();
if (Math.Abs(busStop.x-xPosition.x)+Math.Abs(busStop.y-xPosition.y)<=walkingDistance)
sum++;
}
return sum;
}

public class Position
{
public int x=0;
public int y=0;
}

dotNet的程序员 2005-12-22 14:53 发表评论

re: Google�~�程挑战�?50分题目及(qi��ng)�{�案

FinalFantasy — Thu, 15 Dec 2005 08:33:00 GMT

呵呵�Q�谢谢Spirit了。其实刚开始的时候我也是惛_��用递归的，但是觉得用这�U�方法好理解一些，所以就造这�U�思�\写出来了..........

FinalFantasy 2005-12-15 16:33 发表评论

re: Google�~�程挑战�?50分题目及(qi��ng)�{�案

Spirit — Thu, 15 Dec 2005 05:32:00 GMT

一�U�递归解法�Q?

public class SkipStones {
private String water = "...X...";

public int maxDistance(String water) {
this.water = water;
int max = 0;
int sum = 0;
for (int initial = 1; initial < water.length() + 1; initial++) {
sum = bounce(0, initial);
max = (sum > max ? sum : max);
}
return max;
}

private int bounce(int startDistance, int bounceDistance) {
if (bounceDistance == 0)
return startDistance;

if ((startDistance + bounceDistance) > water.length())
return -1;

if (water.charAt(startDistance + bounceDistance - 1) == 'X')
return startDistance;

return bounce(startDistance + bounceDistance, bounceDistance / 2);
}

public static void main(String[] args) {
SkipStones skipStones = new SkipStones();
System.out.println(skipStones.maxDistance(args[0]));
}
}

Spirit 2005-12-15 13:32 发表评论

re: Google�~�程挑战�?50分题目及(qi��ng)�{�案

FinalFantasy — Thu, 15 Dec 2005 03:28:00 GMT

呵呵�Q�很不��y的是比赛那天我发烧了�Q�进��d��把题目给拷了出来�Q�（如果那天我没生病的话�Q�我想通过入围赛是不成问题的，不过其它的就不敢说了�Q�毕竟谁也不知道google的下一步会(x��)是怎样的：(x��)�Q?img src ="http://www.aygfsteel.com/FinalFantasy/aggbug/23997.html" width = "1" height = "1" />

FinalFantasy 2005-12-15 11:28 发表评论