《Java面试试题》找出连续数组中相同的两个数值

如题：
Find out duplicate number between 1 to N numbers.
描述：
You have got a range of numbers between 1 to N, where one of the number is repeated. You need to write a program to find out the duplicate number.
代码：

package test;import java.util.ArrayList;import java.util.Iterator;/* *  You have got a range of numbers between 1 to N, where one of the number is repeated.  *  You need to write a program to find out the duplicate number. */public class DuplicateNumber {    /*     *  1, 2, 2, 3, 4 ---- Duplicated List     *  Sum(Duplicated List) - 4*5/2 = 2     */    public int findDuplicatNumber(ArrayList<Integer> numberList){        // If there exists a duplicated number, the biggest number in the list is size-1;        int biggestNum = numberList.size() - 1;        // 2S = n*(n+1)        int duplicatedNum = sumList(numberList) - biggestNum*(biggestNum+1)/2;        return duplicatedNum;    }    // get the summation of the list    public int sumList(ArrayList<Integer> numberList){        int sum = 0;        for (int num : numberList) {            sum += num;        }        return sum;    }    public static void main(String[] args) {        ArrayList<Integer> numberArrayList = new ArrayList<Integer>();        for (int i = 1; i <= 20; i++) {            numberArrayList.add(i);        }        numberArrayList.add(10);        DuplicateNumber dumTest = new DuplicateNumber();        System.out.println(dumTest.findDuplicatNumber(numberArrayList));    }}

使用数组的前N项和公式，然后根据连续的特性，利用差值求出数组中重复的数据。
要点
1. Java数组的ArrayList使用方法
2. 数组求和的方法