求数组的所有排列组合数

Problem Description
How many nondecreasing subsequences can you find in the sequence S = {s1, s2, s3, …., sn} ? For example, we assume that S = {1, 2, 3}, and you can find seven nondecreasing subsequences, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}.
Input
The input consists of multiple test cases. Each case begins with a line containing a positive integer n that is the length of the sequence S, the next line contains n integers {s1, s2, s3, …., sn}, 1 <= n <= 100000, 0 <= si <= 2^31.
Output
For each test case, output one line containing the number of nondecreasing subsequences you can find from the sequence S, the answer should % 1000000007.
Sample Input
3
1 2 3
Sample Output
7

代码：

static List<String> list = new ArrayList<String>();    /**      * 求a中count个数的组合      * @param a          存放组合的数           * @param restCount  当前还需多少个数构成一个组合      * @param count      在a中挑选出count个数      * @param b            存放当前组合      * @param pos        指向a中的元素,      */      public static void combination(int []a,int restCount,int count,int[] b,int pos){          if(restCount <= 0 || count > a.length){              return;          }          for(int i = pos;i >= restCount - 1;i-- ){              StringBuilder sb = new StringBuilder();            b[restCount - 1] = a[i];              if(restCount - 1 == 0){                  for(int j:b){                      sb.append(j);                 }                if (!list.contains(sb.toString())) {                    list.add(sb.toString());                }            }              combination(a, restCount-1, count, b, i-1);          }      }      public static void main(String[] args) {        int n ;        Scanner in = new Scanner(System.in);        n = in.nextInt();        int[] s = new int [n];        for (int j = 0; j < s.length; j++) {            s[j] = in.nextInt();        }        for (int i = 0; i < s.length; i++) {            int[] b = new int[i+1];            combination(s, i+1, i+1, b, s.length - 1);        }        System.out.println(list.size());    }