03
04/2015
[LeetCode] Letter Combinations of a Phone Number
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"].
Note:
Although the above answer is in lexicographical order, your answer could be in any order you want.
解题思路：
回溯法。通过递归进行回溯。代码如下。这里有个技巧，通过数组来记录每个数字可能的字母。另外，传递参数时，以字符指针指向当前扫描字符，传输效率更快。下面的代码在LeetCode的效率仅为2ms。
class Solution {
public:
vector<string> letterCombinations(string digits) {
string letters[] = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
vector<string> result;
char * c = (char*)digits.c_str();
getLetters(result, "", c, letters);
return result;
}
void getLetters(vector<string>& result, string s, char* c, string letters[]){
if(*c == '\0'){
if(s!="")
result.push_back(s);
}else{
string letter = letters[*c - '0'];
int len = letter.length();
for(int i=0;i<len; i++){
getLetters(result, s + letter[i], c+1, letters);
}
}
}
};二次刷题（2015-08-18）
class Solution {
public:
vector<string> letterCombinations(string digits) {
string letters[] = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
vector<string> result;
helper(digits, 0, letters, "", result);
return result;
}
void helper(string& s, int n, string* letters, string item, vector<string>& result){
int len = s.length();
if(n >= len){
if(item!=""){
result.push_back(item);
}
return;
}
for(int i = 0; i<letters[s[n] - '0'].length(); i++){
helper(s, n+1, letters, item + letters[s[n]-'0'][i], result);
}
}
};
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