[LeetCode] 17. 电话号码的字母组合

标签：python   commons   数字   letters   turn   array   pad   rac   help   

题目描述:https://leetcode-cn.com/problems/letter-combinations-of-a-phone-number/

题目描述:

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"].

思路:

思路1:

递归

递归过程中记录组合.

思路2:

迭代,类似BFS,每次更新一个字母

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代码:

思路1:

python

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

java

class Solution {
    public List<String> letterCombinations(String digits) {
        if (digits == null || digits.length() == 0) {
            return new ArrayList();
        }
        Map<Character, String> map = new HashMap<Character, String>();
        map.put('2', "abc");
        map.put('3', "def");
        map.put('4', "ghi");
        map.put('5', "jkl");
        map.put('6', "mno");
        map.put('7', "pqrs");
        map.put('8', "tuv");
        map.put('9', "wxyz");
        List<String> res = new LinkedList<String>();
        helper("", digits, 0, res, map);
        return res;

    }

    private void helper(String s, String digits, int i, List<String> res, Map<Character, String> map) {
        if (i == digits.length()) {
            res.add(s);
            return;
        }
        String letters = map.get(digits.charAt(i));
        for (int j = 0; j < letters.length(); j++){
            helper(s+letters.charAt(j),digits,i+1,res,map);
        }
        
    }
}

思路2:

python

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 []
        res = [""]
        for num in digits:
            next_res = []
            for alp in lookup[num]:
                for tmp in res:
                    next_res.append(tmp + alp)
            res = next_res
        return res

c++

vector<string> letterCombinations(string digits) {
    vector<string> res;
    string charmap[10] = {"0", "1", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
    res.push_back("");
    for (int i = 0; i < digits.size(); i++)
    {
        vector<string> tempres;
        string chars = charmap[digits[i] - '0'];
        for (int c = 0; c < chars.size();c++)
            for (int j = 0; j < res.size();j++)
                tempres.push_back(res[j]+chars[c]);
        res = tempres;
    }
    return res;
}

[LeetCode] 17. 电话号码的字母组合

标签：python   commons   数字   letters   turn   array   pad   rac   help   

原文地址：https://www.cnblogs.com/powercai/p/10763521.html