題目：

解法：

Simple and efficient iterative solution.

Explanation with sample input "123"

Initial state:

• result = {""}

Stage 1 for number "1":

• result has {""}
• candiate is "abc"
• generate three strings "" + "a", ""+"b", ""+"c" and put into tmp,
  tmp = {"a", "b","c"}
• swap result and tmp (swap does not take memory copy)
• Now result has {"a", "b", "c"}

Stage 2 for number "2":

• result has {"a", "b", "c"}
• candidate is "def"
• generate nine strings and put into tmp,
  "a" + "d", "a"+"e", "a"+"f",
  "b" + "d", "b"+"e", "b"+"f",
  "c" + "d", "c"+"e", "c"+"f"
• so tmp has {"ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf" }
• swap result and tmp
• Now result has {"ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf" }

Stage 3 for number "3":

• result has {"ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf" }
• candidate is "ghi"
• generate 27 strings and put into tmp,
• add "g" for each of "ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"
• add "h" for each of "ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"
• add "h" for each of "ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"
• so, tmp has
  {"adg", "aeg", "afg", "bdg", "beg", "bfg", "cdg", "ceg", "cfg"
  "adh", "aeh", "afh", "bdh", "beh", "bfh", "cdh", "ceh", "cfh"
  "adi", "aei", "afi", "bdi", "bei", "bfi", "cdi", "cei", "cfi" }
• swap result and tmp
• Now result has
  {"adg", "aeg", "afg", "bdg", "beg", "bfg", "cdg", "ceg", "cfg"
  "adh", "aeh", "afh", "bdh", "beh", "bfh", "cdh", "ceh", "cfh"
  "adi", "aei", "afi", "bdi", "bei", "bfi", "cdi", "cei", "cfi" }

Finally, return result.

class Solution { public:vector<string> letterCombinations(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 casefor(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;} };

總結

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