java JDK里面容器类的排序算法使用的主要是插入排序和归并排序，可能不同版本的实现有所不同，关键代码如下：
代码如下：

/**
     * Performs a sort on the section of the array between the given indices
     * using a mergesort with exponential search algorithm (in which the merge
     * is performed by exponential search). n*log(n) performance is guaranteed
     * and in the average case it will be faster then any mergesort in which the
     * merge is performed by linear search.
     *
     * @param in -
     *            the array for sorting.
     * @param out -
     *            the result, sorted array.
     * @param start
     *            the start index
     * @param end
     *            the end index + 1
     */
    @SuppressWarnings("unchecked")
    private static void mergeSort(Object[] in, Object[] out, int start,
            int end) {
        int len = end - start;
        // use insertion sort for small arrays
        if (len <= SIMPLE_LENGTH) {
            for (int i = start + 1; i < end; i++) {
                Comparable<Object> current = (Comparable<Object>) out[i];
                Object prev = out[i - 1];
                if (current.compareTo(prev) < 0) {
                    int j = i;
                    do {
                        out[j--] = prev;
                    } while (j > start
                            && current.compareTo(prev = out[j - 1]) < 0);
                    out[j] = current;
                }
            }
            return;
        }
        int med = (end + start) >>> 1;
        mergeSort(out, in, start, med);
        mergeSort(out, in, med, end);

        // merging

        // if arrays are already sorted - no merge
        if (((Comparable<Object>) in[med - 1]).compareTo(in[med]) <= 0) {
            System.arraycopy(in, start, out, start, len);
            return;
        }
        int r = med, i = start;

        // use merging with exponential search
        do {
            Comparable<Object> fromVal = (Comparable<Object>) in[start];
            Comparable<Object> rVal = (Comparable<Object>) in[r];
            if (fromVal.compareTo(rVal) <= 0) {
                int l_1 = find(in, rVal, -1, start + 1, med - 1);
                int toCopy = l_1 - start + 1;
                System.arraycopy(in, start, out, i, toCopy);
                i += toCopy;
                out[i++] = rVal;
                r++;
                start = l_1 + 1;
            } else {
                int r_1 = find(in, fromVal, 0, r + 1, end - 1);
                int toCopy = r_1 - r + 1;
                System.arraycopy(in, r, out, i, toCopy);
                i += toCopy;
                out[i++] = fromVal;
                start++;
                r = r_1 + 1;
            }
        } while ((end - r) > 0 && (med - start) > 0);

        // copy rest of array
        if ((end - r) <= 0) {
            System.arraycopy(in, start, out, i, med - start);
        } else {
            System.arraycopy(in, r, out, i, end - r);
        }
    }

看到编程珠玑上有一个很有趣的排序算法-位图法其思想是用1位来表示[0~n-1]中的整数是否存在。1表示存在，0表示不存在。即将正整数映射到bit集合中，每一个bit代表其映射的正整数是否存在。

比如{1，2，3，5，8，13}使用下列集合表示：

  0 1 1 1 0 1 0 0 1 0 0 0 0 1 0 0 0 0 0 0

伪代码如下：

for (i  in  [0~n-1])  bit[i] = 0;
for(i  in [0~n-1])
  if (i in input file)     
    bit[i] = 1

for(i  in [0~n-1])
  if(bit[i] == 1) 
    output i

用java 代码尝试下，效率果然不错：
代码如下：

public class javaUniqueSort {
    public static int[] temp = new int[1000001];
    public static List<Integer> tempList = new ArrayList<Integer>();
    public static int count;

    public static void main(final String[] args) {
        List<Integer> firstNum = new ArrayList<Integer>();
        List<Integer> secondNum = new ArrayList<Integer>();

        for (int i = 1; i <= 1000000; i++) {
            firstNum.add(i);
            secondNum.add(i);
        }

        Collections.shuffle(firstNum);
        Collections.shuffle(secondNum);

        getStartTime();
        Collections.sort(firstNum);
        getEndTime("java sort run time  ");

        getStartTime();
        secondNum = uniqueSort(secondNum);
        getEndTime("uniqueSort run time ");

    }

    public static List<Integer> uniqueSort(final List<Integer> uniqueList) {
        javaUniqueSort.tempList.clear();
        for (int i = 0; i < javaUniqueSort.temp.length; i++) {
            javaUniqueSort.temp[i] = 0;
        }
        for (int i = 0; i < uniqueList.size(); i++) {
            javaUniqueSort.temp[uniqueList.get(i)] = 1;
        }
        for (int i = 0; i < javaUniqueSort.temp.length; i++) {
            if (javaUniqueSort.temp[i] == 1) {
                javaUniqueSort.tempList.add(i);
            }
        }

        return javaUniqueSort.tempList;
    }

    public static void getStartTime() {
        javaShuffle.start = System.nanoTime();
    }

    public static void getEndTime(final String s) {
        javaShuffle.end = System.nanoTime();
        System.out.println(s + ": " + (javaShuffle.end - javaShuffle.start) + "ns");
    }
}

运行时间：

java sort run time  : 1257737334ns
uniqueSort run time : 170228290ns
java sort run time  : 1202749828ns
uniqueSort run time : 169327770ns

如果有重复数据，可以修改下：
代码如下：

public class javaDuplicateSort {
    public static List<Integer> tempList = new ArrayList<Integer>();
    public static int count;

    public static void main(final String[] args) {
        Random random = new Random();
        List<Integer> firstNum = new ArrayList<Integer>();
        List<Integer> secondNum = new ArrayList<Integer>();

        for (int i = 1; i <= 100000; i++) {
            firstNum.add(i);
            secondNum.add(i);
            firstNum.add(random.nextInt(i + 1));
            secondNum.add(random.nextInt(i + 1));
        }
        Collections.shuffle(firstNum);
        Collections.shuffle(secondNum);

        getStartTime();
        Collections.sort(firstNum);
        getEndTime("java sort run time  ");

        getStartTime();
        secondNum = uniqueSort(secondNum);
        getEndTime("uniqueSort run time ");

    }

    public static List<Integer> uniqueSort(final List<Integer> uniqueList) {
        javaDuplicateSort.tempList.clear();
        int[] temp = new int[200002];
        for (int i = 0; i < temp.length; i++) {
            temp[i] = 0;
        }
        for (int i = 0; i < uniqueList.size(); i++) {
            temp[uniqueList.get(i)]++;
        }
        for (int i = 0; i < temp.length; i++) {
            for (int j = temp[i]; j > 0; j--) {
                javaDuplicateSort.tempList.add(i);
            }
        }

        return javaDuplicateSort.tempList;
    }

    public static void getStartTime() {
        javaShuffle.start = System.nanoTime();
    }

    public static void getEndTime(final String s) {
        javaShuffle.end = System.nanoTime();
        System.out.println(s + ": " + (javaShuffle.end - javaShuffle.start) + "ns");
    }
}

这种算法还是有很明显的局限性的，比如说要知道数据中最大的数值，更重要的是数据的疏密程度，比如说最大值为1000000而要数组大小只有100，那么效率会下降的非常明显。。。。。但是，使用位图法进行排序，确实让人眼前一亮。位图法通常是用来存储数据，判断某个数据存不存在或者判断数组是否存在重复 。