题解 | #电话号码的字母组合#

dfs,递归拼接每个字符得到最终的字符串

#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param num string字符串 
# @return string字符串一维数组
#
class Solution:
    def phoneNumber(self , num: str) -> List[str]:
        # write code here
        chars = {"2":"abc", "3":"def", "4":"ghi", "5":"jkl", "6":"mno", "7":"pqrs", "8":"tuv", "9": "wxyz"}
        res= []
        def dfs(s, t):
            nonlocal res
            if len(t) == len(num):
                res.append(t)
                return
            for i in range(s, len(num)):
                for d in chars[num[i]]:
                    dfs(i + 1, t + d)
        dfs(0, "")
        return res