[LeetCode]17. Letter Combinations of a Phone Number
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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"
初始化
result = {""}
第一步:“1”
result = {""}
下一步待处理字符串 "abc"
产生三个操作"" + "a", ""+"b", ""+"c" 将结果插入tmp,
tmp = {"a", "b","c"}
交换result和tmp
result={"a", "b", "c"}
第二步: "2":
result = {"a", "b", "c"}
待处理字符串是 "def"
将产生下面9个操作,
"a" + "d", "a"+"e", "a"+"f",
"b" + "d", "b"+"e", "b"+"f",
"c" + "d", "c"+"e", "c"+"f"
tmp = {"ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf" }
交换result , tmp
result ={"ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf" }
第三步: "3":
result h= {"ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf" }
待处理的字符串是 "ghi"
将产生27个插入到tmp尾部操作,
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"
tmp ={"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" }
交换 result 和 tmp
result ={"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" }
最终返回结果result
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"].
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;}例子:当输入“123”时的解释如下:
初始化
result = {""}
第一步:“1”
result = {""}
下一步待处理字符串 "abc"
产生三个操作"" + "a", ""+"b", ""+"c" 将结果插入tmp,
tmp = {"a", "b","c"}
交换result和tmp
result={"a", "b", "c"}
第二步: "2":
result = {"a", "b", "c"}
待处理字符串是 "def"
将产生下面9个操作,
"a" + "d", "a"+"e", "a"+"f",
"b" + "d", "b"+"e", "b"+"f",
"c" + "d", "c"+"e", "c"+"f"
tmp = {"ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf" }
交换result , tmp
result ={"ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf" }
第三步: "3":
result h= {"ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf" }
待处理的字符串是 "ghi"
将产生27个插入到tmp尾部操作,
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"
tmp ={"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" }
交换 result 和 tmp
result ={"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" }
最终返回结果result
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