Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.
Note:
- All numbers will be positive integers.
- The solution set must not contain duplicate combinations.
Example 1:
Input: k = 3, n = 7 Output: [[1,2,4]]
Example 2:
Input: k = 3, n = 9 Output: [[1,2,6], [1,3,5], [2,3,4]]
backtracking
time: C(m, k) = m! / k!(m-k)!
space: O(k + k * # of ans)
class Solution { public List<List<Integer>> combinationSum3(int k, int n) { List<List<Integer>> res = new ArrayList<>(); backtracking(k, n, 1, new ArrayList<>(), res); return res; } private void backtracking(int k, int n, int start, List<Integer> tmp, List<List<Integer>> res) { if(k == 0) { if(n == 0) res.add(new ArrayList<>(tmp)); } for(int i = start; i <= 9; i++) { if(i > n) break; tmp.add(i); backtracking(k - 1, n - i, i + 1, tmp, res); tmp.remove(tmp.size() - 1); } } }
二刷:
time: O(2 ^ k), space: O(k)
class Solution { public List<List<Integer>> combinationSum3(int k, int n) { List<List<Integer>> res = new ArrayList<>(); dfs(k, n, 1, new ArrayList<>(), res); return res; } public void dfs(int k, int target, int start, List<Integer> list, List<List<Integer>> res) { if(k == 0) { if(target == 0) { res.add(new ArrayList<>(list)); } return; } for(int i = start; i <= 9; i++) { if(i > target) { break; } list.add(i); dfs(k - 1, target - i, i + 1, list, res); list.remove(list.size() - 1); } } }