2014
11-19

# LeetCode-Letter Combinations of a Phone Number[暴力枚举]

### 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.

// LeetCode, Letter Combinations of a Phone Number
// 时间复杂度O(3^n)，空间复杂度O(n)
class Solution {
public:
const vector keyboard { " ", "", "abc", "def", // '0','1','2',...
"ghi", "jkl", "mno", "pqrs", "tuv", "wxyz" };

vector letterCombinations (const string &digits) {
vector result;
dfs(digits, 0, "", result);
return result;
}

void dfs(const string &digits, size_t cur, string path,
vector &result) {
if (cur == digits.size()) {
result.push_back(path);
return;
}
for (auto c : keyboard[digits[cur] - '0']) {
dfs(digits, cur + 1, path + c, result);
}
}
};

// LeetCode, Letter Combinations of a Phone Number
// 时间复杂度O(3^n)，空间复杂度O(1)
class Solution {
public:
const vector keyboard { " ", "", "abc", "def", // '0','1','2',...
"ghi", "jkl", "mno", "pqrs", "tuv", "wxyz" };

vector letterCombinations (const string &digits) {
vector result(1, "");
for (auto d : digits) {
const size_t n = result.size();
const size_t m = keyboard[d - '0'].size();

result.resize(n * m);
for (size_t i = 0; i < m; ++i)
copy(result.begin(), result.begin() + n, result.begin() + n * i);

for (size_t i = 0; i < m; ++i) {
auto begin = result.begin();
for_each(begin + n * i, begin + n * (i+1), [&](string &s) {
s += keyboard[d - '0'][i];
});
}
}
return result;
}
};