leetcode 140: Letter Combinations of a Phone Number

最新推荐文章于 2025-05-25 20:50:12 发布

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最新推荐文章于 2025-05-25 20:50:12 发布

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本文介绍了一种算法，用于根据电话号码生成所有可能的字母组合。包括递归和迭代两种实现方式，适用于理解数字到字母映射的应用场景。

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Letter Combinations of a Phone Number

Total Accepted: 10538 Total Submissions: 40738

Given a digit string, return all possible letter combinations that the number could represent.

A mapping of digit to letters (just like on the telephone buttons) is given below.

Input:Digit string "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

Note:
Although the above answer is in lexicographical order, your answer could be in any order you want.

2, recursive

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        vector<string> res;
        string temp;
        
        rec(res, temp, digits, 0);
        return res;
    }
    
private:
    void rec(vector<string> &res, string &temp, const string &digits, unsigned int level) {
        if(level == digits.length()) {
            res.push_back(temp);
            return;
        }          
        string str = getMapChars(digits[level]);
        for(int i=0; i< str.length(); i++) {
            temp.push_back( str[i] );
            rec(res, temp, digits, level+1);
            temp.pop_back();
        }
    }

    string getMapChars(char digit) {
        if(digit<'2' || digit >'9') return "";
        string arr[] = {"abc","def","ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"}; 
        return arr[digit - '0' - 2];
    }
};

2. iterative java

public class Solution {
    public List<String> letterCombinations(String digits) {
        //init check 
        List<String> res = new ArrayList<String>();
        if(digits==null || digits.length()==0) return res; 
        res.add("");
        
        String[] map = {"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        
        for(int i=0; i<digits.length(); i++) {
            int index = digits.charAt(i) - '2' ;            
            String s = map[index]; //check valid number
            int len = res.size();
            for(int j=0; j<len; j++) {
                String str = res.get(j);
                res.set(j, str + s.charAt(0));
                for(int m=1; m<s.length(); m++) {
                    res.add(str + s.charAt(m));
                }
            }
        }
        
        return res;
    }
}

3, iterative c++

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        const string letters[] = {"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};

        vector<string> ret(1, "");
        for (int i = 0; i < digits.size(); ++i) {
            for (int j = ret.size() - 1; j >= 0; --j) {
                const string &s = letters[digits[i] - '2'];
                for (int k = s.size() - 1; k >= 0; --k) {
                    if (k)
                        ret.push_back(ret[j] + s[k]);
                    else
                        ret[j] += s[k];
                }
            }
        }

        return ret;
    }
};