Leetcode Letter Combinations of a Phone Number DFS

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https://leetcode.com/problems/letter-combinations-of-a-phone-number/

有一个trick 就是如果是输入"" 这个时候返回空

其他就是代码写的简洁的问题了

#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <iostream>#include <vector>using namespace std;const int SIZE = 10+3;class Solution {public:    inline string getString(int num) {        switch (num) {        case 0 :            return " ";        case 1:            return "";        case 2:            return "abc";        case 3:            return "def";        case 4:            return "ghi";        case 5:            return "jkl";        case 6:            return "mno";        case 7:            return "pqrs";        case 8:            return "tuv";        case 9:            return "wxyz";        }        return "";    }    void dfs(string digits, int ptr, string now) {        if(ptr >= digits.size() && now.size()==digits.size()){            ans.push_back(now);            return;        }        for(int i=ptr;i<digits.size();i++) {            string str = getString(digits[i]-'0');            for(int j=0;j<str.size();j++) {                dfs( digits, i+1, now+str[j] );            }        }    }    vector<string> letterCombinations(string digits) {        ans.clear();        if(digits.size() == 0)return ans;   //WA了一次        dfs(digits, 0, "");        return ans;    }private:    vector <string> ans;};int main() {    Solution s;    int n;    string digits;    while(cin >> digits) {        vector<string> ans = s.letterCombinations(digits);        for(int i=0;i<ans.size();i++) {            cout << ans[i] << ", ";        }        cout << endl;    }    return 0;}

一个漂亮的代码

#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <iostream>#include <vector>using namespace std;class Solution {public:    vector<string> letterCombinations(string digits) {        mapping = {"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};        vector<string> ret;        if(digits.size() == 0)return ret;        letterCombinationsCore(digits, "", ret);        return ret;    }    void letterCombinationsCore(string digits, string sub, vector<string> &ret) {        if (sub.size() == digits.size()) {            ret.push_back(sub);        } else {            int i = digits[sub.size()] - '0';            for (int j = 0; j < mapping[i - 2].size(); ++j) {                sub.push_back(mapping[i - 2][j]);                letterCombinationsCore(digits, sub, ret);                sub.pop_back();            }        }    }    vector<string> mapping;};

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