216. 组合总和 III

class Solution {
    private List ls;
    private int n;
    private LinkedList path;
    private int k;
    private int[] arr = new int[] { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
    public List> combinationSum3(int k, int n) {
        path = new LinkedList();
        ls = new LinkedList();
        this.n = n;
        this.k = k;
        dfs(0, 0, 0);
        return ls;
    }
    private void dfs(int i, int num, int count) {
        if (i > 9 || num > k || count > n)
            return;
        if (num == k && count == n) {         //上面不能等于>= 9就跳出 不然很多需要9才能达到 num==k && count==n
            ls.add(path.clone());
            return;
        }
        for (int x = i; x < 9; x++) {
            path.add(arr[x]);
            dfs(x + 1, num + 1, count + arr[x]);
            path.removeLast();
        }
    }
}

倒着遍历可以减少一些参数的设定

class Solution {
    private List ls;
    private LinkedList path;
    private int[] arr = new int[] { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
    public List> combinationSum3(int k, int n) {
        path = new LinkedList();
        ls = new LinkedList();
        dfs(8, n, k);
        return ls;
    }
    private void dfs(int k, int n, int num) {
        if ( n < 0||num < 0)
            return;
        if (num == 0 && n == 0) {
            ls.add(path.clone());
            return;
        }
        for (int x = k; x >=0 ; x--) {
            path.add(arr[x]);
            dfs(x-1,n- arr[x],num-1);
            path.removeLast();
        }
    }
}