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If already sorted, using binary searchbinary search: using idx as search range, h can be obtained by len - idx, then check if mid > h or mid < h or mid = hO(n) solution using hashmap count each number's apperance
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- backtracking public void backtracking(List<List<Integer>> res, List<Integer> tmp, int num, int tmpSum, int k, int n) { if (tmp.size() == k && tmpSum == n) { res.add(new ArrayList<Integer>(tmp)); return; } for (int i = num; i <= 9; i++) { if (tmpSum + i <= n) { tmp.add(i); backtracking(res, tmp, i + 1, tmpSum + i, k, n); tmp.remove(tmp.size() - 1); } } }