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简单DFS题目
代码:
1 vector<string> res; 2 3 void dfs(string &digits, vector<string> &i2s, string ans, int pos) { 4 if (pos == digits.length()) { 5 res.push_back(ans); 6 return; 7 } 8 9 for (auto c : i2s[digits[pos] - ‘0‘]) 10 dfs(digits, i2s, ans + c, pos + 1); 11 } 12 13 vector<string> letterCombinations(string digits) { 14 vector<string> i2s {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"}; 15 dfs(digits, i2s, "", 0); 16 return res; 17 }
Leetcode#17 Letter Combinations of a Phone Number
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原文地址:http://www.cnblogs.com/boring09/p/4262510.html