LeetCode--017--电话号码的字母组合（java and python）

给定一个仅包含数字 2-9 的字符串，返回所有它能表示的字母组合。

给出数字到字母的映射如下（与电话按键相同）。注意 1 不对应任何字母。

示例:

输入："23"
输出：["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
转自：https://blog.csdn.net/xushiyu1996818/article/details/84334799

class Solution {
    HashMap<Character,char[]> map = new HashMap<>();
    List<String> result = new ArrayList<>();
    public List<String> letterCombinations(String digits) {
        int length = digits.length();
        if(length == 0){
            return result;
        }
        map.put('2',new char[]{'a','b','c'});        
        map.put('3', new char[]{'d','e','f'});
        map.put('4', new char[]{'g','h','i'});
        map.put('5', new char[]{'j','k','l'});
        map.put('6', new char[]{'m','n','o'});
        map.put('7', new char[]{'p','q','r','s'});
        map.put('8', new char[]{'t','u','v'});
        map.put('9', new char[]{'w','x','y','z'});
        combine("",digits);
        return result;
    }
    public void combine(String nowStr,String remain){
        int length = remain.length();
        if(length == 0){
            result.add(nowStr);
            return;
        }
        Character now = remain.charAt(0);
        for(char nowChar:map.get(now)){
            combine(nowStr+nowChar,remain.substring(1));
        }
    }
    
}

2019-04-15 22:48:47

23  => ["ad","bd","cd","ae","be","ce","af","bf","cf"]

234=> ["adg","bdg","cdg","aeg","beg","ceg","afg","bfg","cfg","adh"。。。。。。]

class Solution:
    def letterCombinations(self, digits):
        self.dict = {"2":"abc", "3":"def", "4":"ghi", "5":"jkl", "6":"mno", "7":"pqrs","8":"tuv","9":"wxyz"}
        if digits == "":
            return []
        result = [""]
        for digit in digits:
            strs = self.dict[digit]
            curResult = []
            for char in strs:
                for res in result:
                    curResult.append(res+char)
            result = curResult
        return result

2019-11-28 14:13:09