17. Letter Combinations of a Phone Number
题目
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.
解析
- 可以迭代，即依次读取字符串中的每位数字，然后把数字对应的字母依次加到当前的所有结果中，然后进入下一次迭代。也可以用递归来解，思路也类似，就是对于当前已有的字符串，递归剩下的数字串，然后得到结果后加上去。假设输入字符串总共有n个数字，平均每个数字可以代表m个字符，那么时间复杂度是O(m^n)，确切点是输入字符串中每个数字对应字母数量的乘积，即结果的数量，空间复杂度也是一样。时间复杂度：O(m^n)；空间复杂度：O(m^n)
// 17. Letter Combinations of a Phone Number
class Solution_17 {
public:
// map<int, string> num2alp = { { 2, "abc" }, { 3, "def" }, { 4, "ghi" }, { 5, "jkl" }, { 6, "mno" }, { 7, "pqrs" }, { 8, "tuv" }, { 9, "wxyz" } };
/*static const*/ vector<string> v = { "", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz" };
void help(string digits,vector<string> &vec,string &temp,int index,int n)
{
if (index==n)
{
vec.push_back(temp);
return;
}
string str = v[digits[index]-'0'];
for (int i = 0; i < str.size();i++)
{
temp.push_back(str[i]);
help(digits, vec,temp, index + 1,n);
temp.pop_back();
}
return;
}
// 回溯递归实现
vector<string> letterCombinations(string digits) { //23
vector<string> vec;
string temp;
if (digits.size()==0)
{
vec.push_back(""); //初始化一个空
return vec;
}
help(digits,vec,temp,0,digits.size());
return vec;
}
// 迭代实现
vector<string> letterCombinations_ref(string digits) {
vector<string> result;
if (digits.empty()) return vector<string>();
static const vector<string> v = { "", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz" };
result.push_back(""); // add a seed for the initial case
for (int i = 0; i < digits.size(); ++i) {
int num = digits[i] - '0';
if (num < 0 || num > 9) break;
const string& candidate = v[num];
if (candidate.empty()) continue;
vector<string> tmp;
for (int j = 0; j < candidate.size(); ++j) {
for (int k = 0; k < result.size(); ++k) {
tmp.push_back(result[k] + candidate[j]);
}
}
result.swap(tmp); //
}
return result;
}
};
题目来源
