【Medium】17. Letter Combinations of a Phone Number-优快云博客

本文链接：https://blog.youkuaiyun.com/shirleyleychen/article/details/52559697

本文介绍了一种算法，该算法能够根据电话按键上数字到字母的映射，输入一个数字字符串并返回所有可能的字母组合。通过使用深度优先搜索(DFS)来遍历所有可能的组合。

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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"].

中心思想：

先将每一个digit转换成的string存在一个vector中，然后用dfs算出所有组合。

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        vector<string> result;
        vector<string> tmp;
        int n = digits.size();
        if (n == 0)
            return result;
        
        for (int i = 0; i < n; i++){
            switch (digits[i]){
                case '2':
                    tmp.push_back("abc");
                    break;
                case '3':
                    tmp.push_back("def");
                    break;
                case '4':
                    tmp.push_back("ghi");
                    break;
                case '5':
                    tmp.push_back("jkl");
                    break;
                case '6':
                    tmp.push_back("mno");
                    break;
                case '7':
                    tmp.push_back("pqrs");
                    break;
                case '8':
                    tmp.push_back("tuv");
                    break;
                case '9':
                    tmp.push_back("wxyz");
                    break;
                default:
                    return result;
            }
        }
        string comb;
        dfs(tmp, 0, comb, result);
        
        return result;
        
    }
    
    void dfs(vector<string>& tmp, int pos, string& comb, vector<string>& result){
        //recursion先写base case, end case
        if (pos >= tmp.size()){
            result.push_back(comb);
            return;
        }
        int m = tmp[pos].size();
        string s = tmp[pos];
        
        for (int i = 0; i < m; i++){
            comb.push_back(s[i]);
            dfs(tmp, pos+1, comb, result);
            comb.pop_back();
        }
    }
};