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  • LeetCode--017--*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"。。。。。。]

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

    2019-11-28 14:13:09

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  • 原文地址:https://www.cnblogs.com/NPC-assange/p/10713927.html
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