216. 组合总和 III

必须要加上if(n < 0)return;之类基础的剪枝优化才能通过。

class Solution {
    LinkedList<Integer> path = new LinkedList<>();
    List<List<Integer>> res = new LinkedList<>();
    public List<List<Integer>> combinationSum3(int k, int n) {
        backtracking(k, n, 1);
        return res;
    }
    private void backtracking(int k, int n, int stIdx){
        if(n < 0)return;
        if(path.size() > k)return;

        if(path.size() == k && n == 0){
            res.add(new LinkedList<>(path));
            return;
        }

        for(int i = stIdx; i <= 9 - (k - path.size()) + 1; i++){
            path.add(i);
            backtracking(k, n - i, i + 1);
            path.removeLast();
        }
    }
}

#17. 电话号码的字母组合

class Solution {
    String[] mapping = new String[]{"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
    List<String> res = new LinkedList<String>();

    public List<String> letterCombinations(String digits) {
        if(digits.isEmpty())return res;

        backtrack(digits, 0, new StringBuilder());
        return res;
    }
    private void backtrack(String digits, int stIdx, StringBuilder sb){
        if(sb.length() == digits.length()){
            res.add(sb.toString());
            return;
        }

        for(int i = stIdx; i < digits.length(); i++){ // 遍历digits的每一个数字
            int digit = digits.charAt(i) - '0';
            for(char c : mapping[digit].toCharArray()){ // 每个数字对应字母的回溯
                sb.append(c);
                backtrack(digits, i + 1, sb);
                sb.deleteCharAt(sb.length() - 1);
            }
        }
    }
}