LintCode 90. k数和 II

Your title here...Given n unique integers, number k (1<=k<=n) and target.

Find all possible k integers where their sum is target.

public class Solution {
    /*
     * @param A: an integer array
     * @param k: a postive integer <= length(A)
     * @param targer: an integer
     * @return: A list of lists of integer
     */
    public List<List<Integer>> kSumII(int[] A, int k, int target) {
        // write your code here
        List<List<Integer>> resultList = new ArrayList<>();
        if (A == null || A.length == 0 || k == 0 || k > A.length) {
            List<Integer> childList = new ArrayList<>();
            resultList.add(childList);
            return resultList;
        }
        if (k == 1) {
            for (int i = 0; i < A.length; i++) {
                if (target == A[i]) {
                    List<Integer> childList = new ArrayList<>();
                    childList.add(A[i]);
                    resultList.add(childList);
                    break;
                }
            }
            return resultList;
        } else {
            while (A.length >= 1) {
                int num = A[0];
                A = Arrays.copyOfRange(A, 1, A.length);
                List<List<Integer>> list = kSumII(A, k - 1, target - num);
                for (List<Integer> childList : list) {
                    if (!childList.isEmpty()) {
                        childList.add(num);
                        resultList.add(childList);
                    }
                }
            }
            return resultList;
        }
    }
}