LeetCode 17: Letter Combinations of a Phone Number-优快云博客

本文探讨了使用递归和循环两种方法解决电话号码组合问题，具体包括字母组合生成、数字到字母映射及算法实现细节。

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Letter Combinations of a Phone Number

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"].

解题思路

思路一：使用递归：

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        int dlen = digits.size();
        vector<string> result;

        if (dlen == 0 || digits[0] == '1' || digits[0] == '0') {
            return result;
        }

        // 递归
        vector<string> temp = letterCombinations(digits.substr(1, dlen - 1));

        int tempLen = temp.size();
        // 当前数字对应的字符的起始位置
        int i = digits[0] < '8' ? (digits[0] - '2') * 3 : (digits[0] - '2') * 3 + 1; 
        // 当前数字对应的字符个数
        int len = (digits[0] == '9' || digits[0] == '7') ? 4 : 3;
        if (tempLen == 0) {
            if (dlen == 1) {
                // 字符串中不包含 '0' 或 '1'
                for (int k = 0; k < len; ++k) {
                    string str(1, 'a' + i + k);
                    result.push_back(str);
                }
            }
        }
        else {
            for (int k = 0; k < len; ++k) {
                string str(1, 'a' + i + k);
                for (int j = 0; j < tempLen; ++j) {
                    // 将当前字符拼接到每一个递归返回的字符串中
                    result.push_back(str + temp[j]);
                }
            }
        }
        return result;
    }
};

思路二：使用循环：

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        vector<string> ret;
        if(digits == "")
            return ret;

        ret.push_back("");

        vector<string> dict(10); //0~9
        dict[2] = "abc";
        dict[3] = "def";
        dict[4] = "ghi";
        dict[5] = "jkl";
        dict[6] = "mno";
        dict[7] = "pqrs";
        dict[8] = "tuv";
        dict[9] = "wxyz";

        for(int i = 0; i < digits.size(); i ++)
        {
            int size = ret.size();
            for(int j = 0; j < size; j ++)
            {
                string cur = ret[0];
                ret.erase(ret.begin());
                for(int k = 0; k < dict[digits[i]-'0'].size(); k ++)
                {
                    ret.push_back(cur + dict[digits[i]-'0'][k]);
                }
            }
        }
        return ret;
    }
};