抢答题！！！

Question

例如一个整型数组　int temp[] = {2,2,1,2,3,3,2}
其中２这个元素的个数大于了数组长度的一半，则称２为这个数组的“持重点”；
用程序算出给定这个数组的持重点是几，并返回它的所有下标；

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[b]问题补充：[/b]
怎么就会完全不明白呢？就拿这个数组来说，现在我们知道２是持重点，假如在不知道的情况下，让你求出来谁是持重点，并返回这个持重点的所有下标。
谁把程序写出来，我把仅有的积分全贡出去。

Answer 1

[code="java"]import java.util.ArrayList;
import java.util.HashMap;
import java.util.Iterator;
import java.util.List;
import java.util.Map;

public class Test {

public static void main(String[] args) {
    int temp[] = {2,2,1,2,3,3,2};
    int chiZhongDian = getChiZhongDian(temp);

    System.out.println("chiZhongDian:" + chiZhongDian);

    List appearList = getIndex(temp , chiZhongDian);

    System.out.println("index in Array:");
    for(Iterator itr = appearList.iterator();itr.hasNext();)
    {
        System.out.print(itr.next() + " ");
    }

}

public static List getIndex(int[] temp, int number)
{
    List appearList = new ArrayList();

    for(int i = 0;i < temp.length;i ++)
    {
        if(temp[i] == number)
        {
            appearList.add(new Integer(i));
        }
    }

    return appearList;
}

public static int getChiZhongDian(int[] temp) {
    int size = temp.length;

    Map chiZhongDianMap = new HashMap();
    for (int i = 0; i < temp.length; i++) {
        if (chiZhongDianMap.get(new Integer(temp[i])) == null) {
            Integer eachCount = new Integer(1);

            chiZhongDianMap.put(new Integer(temp[i]), new Integer(1));
        } else {
            Integer eachCount = (Integer) chiZhongDianMap
                    .get(new Integer(temp[i]));
            eachCount = new Integer(eachCount.intValue() + 1);
            chiZhongDianMap.put(new Integer(temp[i]), eachCount);
        }
    }

    for(Iterator itr = chiZhongDianMap.keySet().iterator();itr.hasNext();)
    {
        Integer number = (Integer)itr.next();
        Integer eachCount = (Integer)chiZhongDianMap.get(number);

        if(eachCount.intValue() > (size / 2))
            return number.intValue();
    }

    throw new RuntimeException("Can not find ChiZhongDian");
}

}[/code]