LeetCode: "Letter Combinations of a Phone Number

一些小问题， “”的输出我的是[], 但是答案却是[""]， 比较难理解，3次过

 1 class Solution {

 2 public:

 3     void dfs(vector<string> &ret, string tmp, map<char, string> refer, string digits, int beg) {

 4         if (beg >= digits.size()) {

 5             ret.push_back(tmp);

 6             return;

 7         }

 8         if (digits[beg] <= '9' && digits[beg] >= '2') {

 9             for (int i = 0; i < refer[digits[beg]].size(); i++) {

10                 string r = refer[digits[beg]];

11                 dfs(ret, tmp+r[i], refer, digits, beg+1);

12             }

13         }

14     }

15     vector<string> letterCombinations(string digits) {

16         // Start typing your C/C++ solution below

17         // DO NOT write int main() function

18         map<char, string> refer;

19         refer['2'] = "abc";

20         refer['3'] = "def";

21         refer['4'] = "ghi";

22         refer['5'] = "jkl";

23         refer['6'] = "mno";

24         refer['7'] = "pqrs";

25         refer['8'] = "tuv";

26         refer['9'] = "wxyz";

27         vector<string> ret;

28         string tmp = "";

29         dfs(ret, tmp, refer, digits, 0);

30         return ret;

31     }

32 };

 加一个iterative的方法

 1 class Solution {

 2 public:

 3     vector<string> letterCombinations(string digits) {

 4         // IMPORTANT: Please reset any member data you declared, as

 5         // the same Solution instance will be reused for each test case.

 6         map<char, string> S;

 7         S['2'] = "abc", S['3'] = "def", S['4'] = "ghi", S['5'] = "jkl";

 8         S['6'] = "mno", S['7'] = "pqrs", S['8'] = "tuv", S['9'] = "wxyz";

 9         queue<string> que;

10         vector<string> res;

11         if (digits.size() == 0) {

12             res.push_back("");

13             return res;

14         }

15         for (int i = 0; i < S[digits[0]].size(); i++) que.push(string("")+S[digits[0]][i]);

16         while (!que.empty()) {

17             string tmp = que.front();

18             que.pop();

19             if (tmp.size() == digits.size()) res.push_back(tmp);

20             else for (int i = 0; i < S[digits[tmp.size()]].size(); i++) que.push(tmp+S[digits[tmp.size()]][i]);

21         }

22         return res;

23     }

24 };

 C#

 1 public class Solution {

 2     public List<string> LetterCombinations(string digits) {

 3         Dictionary<char, string> S = new Dictionary<char, string>();

 4         S.Add('2', "abc"); S.Add('3', "def"); S.Add('4', "ghi"); S.Add('5', "jkl");

 5         S.Add('6', "mno"); S.Add('7', "pqrs"); S.Add('8', "tuv"); S.Add('9', "wxyz");

 6         Queue<string> que = new Queue<string>();

 7         List<string> ans = new List<string>();

 8         if (digits.Length == 0) return ans;

 9         foreach (var s in S[digits[0]]) que.Enqueue("" + (char)s);

10         while (que.Count != 0) {

11             string peek = que.Peek();

12             que.Dequeue();

13             if (peek.Length == digits.Length) ans.Add(peek);

14             else foreach(var s in S[digits[peek.Length]]) que.Enqueue(peek + (char)s);

15         }

16         return ans;

17     }

18 }

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