算法修炼Day25|● 216.组合总和III ● 17.电话号码的字母组合

题目：216. 组合总和 III - 力扣（LeetCode）

思路/想法：

陷入了参考（抄）答案的怪圈。。

代码实现：

class Solution {
    List<List<Integer>> ans = new ArrayList<>();
    LinkedList<Integer> path = new LinkedList<>();

    public List<List<Integer>> combinationSum3(int k, int n) {
        backtracking(n, k, 1, 0);
        return ans;
    }
    private void backtracking(int n, int k, int startIndex, int sum) {
        if (sum > n) {
            return;
        }
        if (path.size() > k) return;
        if (path.size() == k) {
            if (sum == n ) {
                ans.add(new ArrayList<>(path));
            }
            return;
        }
        for (int i = startIndex; i <= 9; i++) {
            path.add(i);
            sum += i;
            backtracking(n, k, i + 1, sum);
            sum -= i;
            path.removeLast();
        }
    }
}

题目：17. 电话号码的字母组合 - 力扣（LeetCode）

代码实现：

class Solution {
    List<String> ans = new ArrayList<>();
    StringBuilder temp = new StringBuilder();

    public List<String> letterCombinations(String digits) {
        if (digits == null || digits.length() == 0) return ans;
        String[] numString = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        backtracking(digits, numString, 0);
        return ans;
    }
    public void backtracking(String digits, String[] numString, int num) {
        if (num == digits.length()) {
            ans.add(temp.toString());
            return;
        }
        String str = numString[digits.charAt(num) - '0'];
        for (int i = 0; i < str.length(); i++) {
            temp.append(str.charAt(i));
            backtracking(digits, numString, num + 1);
            temp.deleteCharAt(temp.length() - 1);
        }
    }
}