Given [1,2,3,4], k=2, target=5. There are 2 solutions:

[1,4] and [2,3], return 2.

 1 /**
 2      * @param A: an integer array.
 3      * @param k: a positive integer (k <= length(A))
 4      * @param target: a integer
 5      * @return an integer
 6      */
 7     public int  kSum1(int A[], int k, int target) {
 8         // write your code here
 9         if (target < 0) {
10             return 0;
11         }
12
13         int len = A.length;
14
15         int[][][] D = new int[len + 1][k + 1][target + 1];
16
17         for (int i = 0; i <= len; i++) {
18             for (int j = 0; j <= k; j++) {
19                 for (int t = 0; t <= target; t++) {
20                     if (j == 0 && t == 0) {
21                         // select 0 number from i to the target: 0
22                         D[i][j][t] = 1;
23                     } else if (!(i == 0 || j == 0 || t == 0)) {
24                         D[i][j][t] = D[i - 1][j][t];
25                         if (t - A[i - 1] >= 0) {
26                             D[i][j][t] += D[i - 1][j - 1][t - A[i - 1]];
27                         }
28                     }
29                 }
30             }
31         }
32
33         return D[len][k][target];
34     }

 1 // 2 dimension
 2     public int  kSum(int A[], int k, int target) {
 3         // write your code here
 4         if (target < 0) {
 5             return 0;
 6         }
 7
 8         int len = A.length;
 9
10         // D[i][j]: k = i, target j, the solution.
11         int[][] D = new int[k + 1][target + 1];
12
13         // only one solution for the empty set.
14         D[0][0] = 1;
15         for (int i = 1; i <= len; i++) {
16             for (int t = target; t > 0; t--) {
17                 for (int j = 1; j <= k; j++) {
18                     if (t - A[i - 1] >= 0) {
19                         D[j][t] += D[j - 1][t - A[i - 1]];
20                     }
21                 }
22             }
23         }
24
25         return D[k][target];
26     }