力扣17. *的字母组合

原题

解法1、2都是大佬写的，题解链接

1、回溯

class Solution:
    def letterCombinations(self, digits: str) -> List[str]:
        if not digits: return []

        phone = {'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 backtrack(conbination,nextdigit):
            if len(nextdigit) == 0:
                res.append(conbination)
            else:
                for letter in phone[nextdigit[0]]:
                    backtrack(conbination + letter,nextdigit[1:])

        res = []
        backtrack('',digits)
        return res

2、队列

class Solution:
    def letterCombinations(self, digits: str) -> List[str]:
        if not digits: return []
        phone = ['abc','def','ghi','jkl','mno','pqrs','tuv','wxyz']
        queue = ['']  # 初始化队列
        for digit in digits:
            for _ in range(len(queue)):
                tmp = queue.pop(0)
                for letter in phone[ord(digit)-50]:# 这里我们不使用 int() 转换字符串，使用ASCII码
                    queue.append(tmp + letter)
        return queue

3、与大佬的相比我的代码就比较烂了

class Solution:
    def letterCombinations(self, digits: str) -> List[str]:
        ans,dic,lens,cnt = [],{},len(digits),0
        dic['2'],dic['3'],dic['4'],dic['5'] = 'abc','def','ghi','jkl'
        dic['6'],dic['7'],dic['8'],dic['9'] = 'mno','pqrs','tuv','wxyz'
        for i in range(lens):
            if i == 0:
                ans += ['']
            for j in range(len(ans)):
                for c in dic[digits[i]]:
                    ans.append(ans[j] + c)
        i = len(ans) - 1
        while i >= 0 and len(ans[i]) == lens:
            i -= 1
        return ans[i + 1:]