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  • [LeetCode] 17. *的字母组合

    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。

    img

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