leetcode17 电话号码的字母组合 ★

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/u012343179/article/details/89875278

思路：暴力法或者递归。

1）暴力法：每次算上一个result字母集合与下一个digit对应字母集合的笛卡尔积

class Solution:
    def letterCombinations(self, digits: str) -> List[str]:
        dict_digit={'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']}
        result=[]
        for d in digits:
            if len(result)==0:
                result=dict_digit[d]
            else:
                temp=[]
                for i in result:
                    for j in dict_digit[d]:
                        temp.append(i+j)
                result=temp
        return result

2）递归法：使用i来表示当前递归到第几个digit，如果i==n，就放入该字母组合，然后回溯。

class Solution:
    def letterCombinations(self, digits: str) -> List[str]:
        lookup = {
            "2":"abc",
            "3":"def",
            "4":"ghi",
            "5":"jkl",
            "6":"mno",
            "7":"pqrs",
            "8":"tuv",
            "9":"wxyz"
        }
        if not digits:
            return []
        n = len(digits)
        res = []
        def helper(i,tmp):
            if i == n:
                res.append(tmp)
                return 
            for alp in lookup[digits[i]]:
                helper(i+1,tmp+alp)
        helper(0,"")
        return res