leetcode | Letter Combinations of a Phone Number

最新推荐文章于 2020-07-24 11:09:14 发布

曲小鑫

最新推荐文章于 2020-07-24 11:09:14 发布

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分类专栏： LeetCode 文章标签： leetcode string

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本文介绍了一种解决电话号码解码问题的算法，通过将数字映射到字母并生成所有可能的字母组合。算法的时间复杂度为O(3^n)，空间复杂度为O(n)，其中n为输入数字串的位数。

问题

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.

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

//时间复杂度 O(3^n)， 空间复杂度O(n) ； n为digit的位数
const string key[] = { "", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz" };
class Solution {
public:
	vector<string> letterCombinations(string digits) {
		vector<string> result;
		if (digits.size() == 0)
			return result;
		string str(digits.size(), ' ');  //为单个字符串分配空间
		PushString(digits, 0, str, result);
		return result;
	}

private:
	void PushString(string& digits, int k, string &str, vector<string>& result) {
		if (k == digits.size()) {
			result.push_back(str);
			return;
		}
		for (int i = 0; i < key[digits[k] - '0'].size(); i++) {
			str[k] = key[digits[k] - '0'][i];  //因为每一个str的长度都等于digits的大小，所以str可以重用（完全覆盖）
			PushString(digits, k + 1, str, result);
		}


	}
};