思路:使用回溯+递归
code:
class Solution
{
public:
/*
* @param digits: A digital string
* @return: all posible letter combinations
*/
vector<string> letterCombinations(string digits)
{
// write your code here
if (digits.size() <= 0)
{
return vector<string>();
}
vector<string> result;
string str;
letterCombinations(digits, str, 0, result);
return result;
}
void letterCombinations(string &digits, string &str, int index, vector<string> &result)
{
if (index == digits.size())
{
result.push_back(str);
// str.clear();
return;
}
string base[] = { "", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz" };
for (int i = 0; i < base[digits[index] - '0'].size(); i++)
{
str += base[digits[index] - '0'][i];
letterCombinations(digits, str, index + 1, result);
str.pop_back(); //使用回溯
}
}
};
转自:https://www.cnblogs.com/libaoquan/p/7381348.html
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