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<List<Integer>> combinationSum3(int k, int n) {
		path = new LinkedList<Integer>();
		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<List<Integer>> combinationSum3(int k, int n) {
		path = new LinkedList<Integer>();
		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();
		}
	}
}