一道500强外企机试题。。。。

[ceasarcn]

传入三个数组，
a={"football","basketball","volleyball"};
b={"adi","rabk","ghy","|","nike","|","adi","rty"};
c={"","","","","",""};

根据a数组元素查找"football",如果在a[0]处找到，就把b数组里第1个"|"前的第一个元素打印c数组长度次。即打印6次System.out.println("adi");
同理，如果在a[1]处找到"football"，就把b数组里第2个"|"前的第一个元素打印c数组长度次。即打印6次System.out.println("nike");

其实我已经写了一个，但是现在总是不好定位到要打印的那个元素。程序后部分有问题的，请各位帮忙指点一下哦。。。先谢了
static void testGenerator(String[] a,String[] b,String[] c)

{

int count = 0,k = 0,end = 0;

for(int i=0;i<a.length;i++)

{

if("ata".equals(a[i]))

{

count = i;

break;

}

}

for(int j=0;j<b.length;j++)

{

if("CC".equals(b[j]))

{

k++;

if(k == count)

{

end = k+1;

break;

}

}

}

for(int n=0;n<c.length;n++)

{

System.out.println(b[end]);

}



}

[WYhack]

学习！！！！

[bocai17]

用正则定位，正则会用不，会用就很简单了

[melody1128]

public void getReturn(String[] a,String[] b,String[] c) {
		String str = "football";
		int times = c.length;
		int index = 2;
		for (int i = 0; i < a.length; i++) {
			if (a[i].equals("football"))
				index = i;
		}
		printStr(getFirStr(index,b),times);
	}

	public String getFirStr(int index,String[] b) {
		String str = null;
		if (index == 0) {
			str = b[0];
		} else if (index == 1) {
			for (int i = 0; i < b.length; i++) {
				if (b[i].equals("|")) {
					str = b[i + 1];
					break;
				}
			}
		}
		
		return str;

	}
	
	public void printStr(String str,int times){
		for(int i=0;i<times;i++){
			System.out.println(str);
		}
	}

[designedIt]

[Quote=引用 4 楼 ceasarcn 的回复:]
感觉需求还是没说清，楼上的方案适合0和1的情况。但是实际上a的元素个数和b的"|"有相对应的关系形式。即如果a有3元素：
a={"football","basketball","volleyball"};
那么b={"ds","fr","|","fde","|","wew","vfr","bgt","|"};
同理如果a有5元素，b中也会出现5次 "|"，所以如果a中第四个出现"football",我们需要把b中第四个"|"前的第一个元素打印c数组长度次。

[/Quote]
我也是菜鸟，我是从你的代码中边学习边试着改的。不要笑话啊。代码好像能运行了，结果好像对的。

public class TestSearch {
   
	static void testGenerator(String[] a,String[] b,String[] c) 

    { 

        int count = -1,k = 0,end = -1; //好像count初值为-1,end初值为-1

        for(int i=0;i <a.length;i++) 

        { 

            if("volleyball".equals(a[i])) 

            { 

                count = i; 

                break; 

            } 
        } 

        if(count==-1){//好像要检查一下
        	System.out.println("数组a中找不到。");return;
        }
        
        if(count==0)
        {
        	end=0;
        }
        else{
        for(int j=0;j <b.length;j++) 

        { 

            if("|".equals(b[j])) //在b中找"|"

            { 

                k++; 

                if(k == count) 

                { 

                    end = j+1; //应该是j+1吧

                    break; 

                } 

            } 

        }//for 
        }//if
        for(int n=0;n <c.length;n++) 

        { 

            System.out.println(b[end]); 

        } 



    }
	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO 自动生成方法存根
		testGenerator(new String[]{"football","basketball","volleyball"},
                new String[]{"ds","fr","|","fde","|","wew","vfr","bgt","|"},
                new String[]{"","","","","",""} );
	}

}

wew
wew
wew
wew
wew
wew

[ceasarcn]

楼上的，十分感谢。。呵呵，确实有一个地方疏忽却一直没看出来，，就是
if(k == count)
{

end = j+1;
break;
}
应该是《j+1》，多谢3楼和5楼的热心帮助。。。:)

[ceasarcn]

感觉需求还是没说清，楼上的方案适合0和1的情况。但是实际上a的元素个数和b的"|"有相对应的关系形式。即如果a有3元素：
a={"football","basketball","volleyball"};
那么b={"ds","fr","|","fde","|","wew","vfr","bgt","|"};
同理如果a有5元素，b中也会出现5次 "|"，所以如果a中第四个出现"football",我们需要把b中第四个"|"前的第一个元素打印c数组长度次。