17. Letter Combinations of a Phone Number

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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

• 17. Letter Combinations of a Phone Number
• LeetCode --- 17. Letter Combinations of a Phone Number