题目链接: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.
1 2(abc) 3(def)
4(ghi) 5(jkl) 6(mno)
7(pqrs) 8(tuv) 9(wxyz)
* 0 #
For example:
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.
这道题的要求是给定一个数字字符串,返回手机键盘上对应的所有可能的字母组合。其中返回的顺序可以是任意的。
这道题的思路比较简单,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)
1. 迭代
1 class Solution
2 {
3 public:
4 vector<string> letterCombinations(string digits) // 迭代
5 {
6 string d[] = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"}, s="";
7 vector<string> v({""});
8 for(int i = 0; i < digits.size(); ++ i)
9 {
10 vector<string> temp;
11 for(int j = 0; j < v.size(); ++ j)
12 for(int k = 0; k < d[digits[i] - ‘0‘].size(); ++ k)
13 temp.push_back(v[j] + d[digits[i] - ‘0‘][k]);
14 v = temp;
15 }
16 return v;
17 }
18 };
2. 递归
1 class Solution
2 {
3 private:
4 vector<string> v;
5 void letterCombinations(string digits, int b, string s, string d[])
6 {
7 if(digits.size() == b)
8 v.push_back(s);
9 else
10 for(int i = 0; i < d[digits[b] - ‘0‘].size(); ++ i)
11 letterCombinations(digits, b + 1, s + d[digits[b] - ‘0‘][i], d);
12 }
13 public:
14 vector<string> letterCombinations(string digits) // 递归
15 {
16 string d[] = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"}, s="";
17 letterCombinations(digits, 0, s, d);
18 return v;
19 }
20 };