Given a string containing digits from 2-9 inclusive, 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. Note that 1 does not map to any letters.
Example:
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| Input: "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.
给定一个仅包含数字 2-9 的字符串，返回所有它能表示的字母组合。
给出数字到字母的映射如下（与电话按键相同）。注意 1 不对应任何字母。
示例:
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| 输入："23"
输出：["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
|
说明:
尽管上面的答案是按字典序排列的，但是你可以任意选择答案输出的顺序。
想法
首先用数组或者map存下来数字和字母的对应关系，然后循环+DFS逐位组合字符即可，实际上这道题和求全排列没有什么区别……
解
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| #include <iostream>
#include <string>
#include <vector>
using namespace std;
class Solution {
public:
char letter[10][5] = {" ", "*", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
vector<string> letterCombinations(string digits)
{
vector<string> ans;
if (digits.size() <= 0) {
return ans;
}
string s;
dfs(ans, s, digits, 0);
return ans;
}
void dfs(vector<string> &ans, string &s, const string &digits, int pos)
{
if (pos >= digits.size()) {
ans.push_back(s);
return;
}
else {
int loc = digits[pos] - '0';
for (int i = 0; letter[loc][i] != '\0'; i++) {
s.push_back(letter[loc][i]);
dfs(ans, s, digits, pos + 1);
s.pop_back();
}
}
}
};
int main(void)
{
Solution s;
vector<string> ans = s.letterCombinations("");
for (string s : ans) {
cout << s << endl;
}
return 0;
}
|
