216. 组合总和 III

1. first idea

I set the root as $n$.

Then, the branches are from $n-1$ to $n-min(9, n)$. For each traversal, the input parameter is a temporal list which stores the number using for subtraction.

As the children for the branch $n-i$, the branches are from $n-(i+1)$ to $n-min(9, n)$.

If any leaves are zeros, their temporal lists are the results.

class Solution:
    def __init__(self):
        self.res_list = []

    def bfs(self, tmp_list, k, n, tmp_sum):
        if k == len(tmp_list):
            if tmp_sum == n:
                self.res_list.append(tmp_list)
            return
        else:
            start_idx = tmp_list[-1] + 1 if tmp_list else 1
            for i in range(start_idx, min(n - tmp_sum, 9) + 1):
                self.bfs(tmp_list + [i], k, n, tmp_sum + i)

    def combinationSum3(self, k: int, n: int) -> List[List[int]]:
        self.bfs([], k, n, 0)
        return self.res_list

Practically, I store the temporal sum value in each node instead of $n-i$.

2. learning time

30min.

17. 电话号码的字母组合

1. first idea

class Solution:
    def __init__(self):
        self.res_list = []
        self.char_dict = {
            "2": ["a", "b", "c"],
            "3": ["d", "e", "f"],
            "4": ["g", "h", "i"],
            "5": ["j", "k", "l"],
            "6": ["m", "n", "o"],
            "7": ["p", "q", "r", "s"],
            "8": ["t", "u", "v"],
            "9": ["w", "x", "y", "z"]
        }

    def dfs(self, tmp_list, left_chars_list):
        if len(left_chars_list) == 0:
            self.res_list.append("".join(tmp_list))
            return
        else:
            ch_list = self.char_dict[left_chars_list[0]]
            for ch in ch_list:
                self.dfs(tmp_list + [ch], left_chars_list[1:])

    def letterCombinations(self, digits: str) -> List[str]:
        digits_list = list(digits)
        self.dfs([], digits_list)
        if self.res_list == [""]:
            return []
        return self.res_list

It is easy.

2. learning time

30min.