String []deptInfo=new String []{"1","产品部","2","实施部","3","销售部","4","市场部","5","人力资源部"};
String []employeeInfo=new String[]{"3","三五","3","互补","4","李婷","1","张三","1","李四","2","核武","5","小虎"};
String a=null;
for (int i = 0; i < deptInfo.length; i+=2) {
    a=deptInfo[i];
    System.out.println(deptInfo[i]+"--------");

    for (int j = 0; j < employeeInfo.length-1; j+=2) {
        if(a.equals(employeeInfo[j])){
            System.out.println(employeeInfo[j+1]+":LLLL");
        }
    }
}

------解决方案--------------------
当下标为奇数时，a的值仍为上一次偶数的赋值，所以重复了。
如：当i=1时，a的值仍为0。
这是我改的
[code=Java][/code]		String []deptInfo=new String []{"1","产品部","2","实施部","3","销售部","4","市场部","5","人力资源部"};
		String []employeeInfo=new String[]{"3","三五","3","互补","4","李婷","1","张三","1","李四","2","核武","5","小虎"};
		String a=null;
		for (int i = 0; i < deptInfo.length; i++) {

		if(i%2==0){

		a=deptInfo[i];
		  
		System.out.println(deptInfo[i+1]+"--------");
		}else
			a=null;//当下标位置为部门名称时清空
		
		for (int j = 0; j < employeeInfo.length; j++) {

		if(j%2==0){
			
			if(a!=null)//判断是否为空
		if(a.equals(employeeInfo[j])){
			

		System.out.println(employeeInfo[j+1]+":LLLL");
		
		}

		}

		}//a=null; 这里清空也行！

		}
	}

------解决方案--------------------
当下标为奇数时，a的值仍为上一次偶数的赋值，所以重复了。
如：当i=1时，a的值仍为0。
这是我改的
Java code
        String []deptInfo=new String []{"1","产品部","2","实施部","3","销售部","4","市场部","5","人力资源部"};
        String []employeeInfo=new String[]{"3","三五","3","互补","4","李婷","1","张三","1","李四","2","核武","5","小虎"};
        String a=null;
        for (int i = 0; i < deptInfo.length; i++) {

        if(i%2==0){

        a=deptInfo[i];
         
        System.out.println(deptInfo[i+1]+"--------");
        }else
            a=null;//当下标位置为部门名称时清空
        
        for (int j = 0; j < employeeInfo.length; j++) {

        if(j%2==0){
            
            if(a!=null)//判断是否为空
        if(a.equals(employeeInfo[j])){
            

        System.out.println(employeeInfo[j+1]+":LLLL");
        
        }

        }

        }//a=null; 这里清空也行！

        }
    }