day25 代码随想录算法训练营 17. 电话号码的字母组合

题目：17. 电话号码的字母组合

我的感悟：

一时间没理解没关系，只要不放弃，就会成长！！！

理解难点：

index是独立集合的起点，需要理解它。
有些东西就是时间的积累

代码难点：

代码示例：

class Solution:
    def __init__(self):
        self.letterMap = [
                "",  # 0
                "",  # 1
                "abc",  # 2
                "def",  # 3
                "ghi",  # 4
                "jkl",  # 5
                "mno",  # 6
                "pqrs",  # 7
                "tuv",  # 8
                "wxyz"  # 9
            ]
    def letterCombinations(self, digits: str) -> List[str]:
        if len(digits)==0:
            return []
        res = []
        self.backtracking(digits,0,[],res)
        return res
    
    def backtracking(self,digits,index,path,res):
        if len(path) == len(digits):
            res.append("".join(path))
            return
        wight = self.letterMap[int(digits[index])] 
        for i in range(len(wight)):
            path.append(wight[i])
            self.backtracking(digits,index+1,path,res)
            path.pop()

补充注释：

class Solution:
    def __init__(self):
        self.letterMap = [
            "",  # 0
            "",  # 1
            "abc",  # 2
            "def",  # 3
            "ghi",  # 4
            "jkl",  # 5
            "mno",  # 6
            "pqrs",  # 7
            "tuv",  # 8
            "wxyz"  # 9
        ]

    def letterCombinations(self, digits: str) -> List[str]:
        # 对特殊做处理
        if len(digits) == 0:
            return []
        # 正常走回溯法
        res = []
        self.backtracking(digits, 0, [], res)
        return res

    def backtracking(self, digits, index, path, res):
        # 1.深度达到要求就跳出循环
        if len(path) == len(digits):
            res.append("".join(path))
            return
        # 2.宽度 . 横向遍历
        wight = self.letterMap[int(digits[index])]  # 宽度,是指每个小集合的宽度 "abc"的宽度,... "wxyz"的宽度
        for i in range(len(wight)):
            path.append(wight[i])
            self.backtracking(digits, index + 1, path, res)  # 这里的index是独立集合的index
            path.pop()