90 - k数和 II

2017.9.29

还是使用的递归的方式

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 targer) {
        // write your code here
		List<List<Integer>> res = new LinkedList<>();
		if(A.length == 0 || k > A.length || k <= 0){
			List<Integer> tmp = new LinkedList<>();
			res.add(tmp);
			return res;
		}
		if(k == 1){
			for(int x : A){
				if(x == targer){
					List<Integer> tmp = new LinkedList<>();
					tmp.add(x);
					res.add(tmp);
				}
			}
			return res;
		}
	    else{
			while(A.length >= 1){
			    int num = A[0];
				A =  Arrays.copyOfRange(A, 1, A.length);
				List<List<Integer>> list = kSumII(A,k-1,targer-num);
				for(List<Integer> listTmp : list){
					if(!listTmp.isEmpty()){
						listTmp.add(num);
						res.add(listTmp);
					}
				}
			}
		}
		return res;
    }
}