day 25: backtracking!

Leetcode 216. Combination Sum III

class Solution:
    def __init__(self):
        self.path = []
        self.res = []
    def combinationSum3(self, k, n):
        nums = [i+1 for i in range(9)]
        self.dfs(nums, n, k)
        return self.res
    def dfs(self, nums, n , k):
        if len(self.path)==k and sum(self.path)==n:
            self.res.append(self.path[:])
            return
        for i in range(len(nums)):
            self.path.append(nums[i])
            self.dfs(nums[i+1:], n, k)
            self.path.pop()

easy~

Leetcode 17. Letter Combinations of a Phone Number

没写出来。。。。。

class Solution:
    def __init__(self):
        self.answers: List[str] = []
        self.answer: str = ''
        self.letter_map = {
            '2': 'abc',
            '3': 'def',
            '4': 'ghi',
            '5': 'jkl',
            '6': 'mno',
            '7': 'pqrs',
            '8': 'tuv',
            '9': 'wxyz'
        }

    def letterCombinations(self, digits: str) -> List[str]:
        self.answers.clear()
        if not digits: return []
        self.backtracking(digits, 0)
        return self.answers
    
    def backtracking(self, digits: str, index: int) -> None:
        # 回溯函数没有返回值
        # Base Case
        if index == len(digits):    # 当遍历穷尽后的下一层时
            self.answers.append(self.answer)
            return 
        # 单层递归逻辑  
        letters: str = self.letter_map[digits[index]]
        for letter in letters:
            self.answer += letter   # 处理
            self.backtracking(digits, index + 1)    # 递归至下一层
            self.answer = self.answer[:-1]  # 回溯