README.md

17. 电话号码的字母组合

English Version

题目描述

给定一个仅包含数字 2-9 的字符串，返回所有它能表示的字母组合。答案可以按 任意顺序 返回。

给出数字到字母的映射如下（与电话按键相同）。注意 1 不对应任何字母。

 

示例 1：

输入：digits = "23"
输出：["ad","ae","af","bd","be","bf","cd","ce","cf"]

示例 2：

输入：digits = ""
输出：[]

示例 3：

输入：digits = "2"
输出：["a","b","c"]

 

提示：

• 0 <= digits.length <= 4
• digits[i] 是范围 ['2', '9'] 的一个数字。

解法

Python3

class Solution:
    def letterCombinations(self, digits: str) -> List[str]:
        n = len(digits)
        if n == 0:
            return []
        chars = ['abc', 'def', 'ghi', 'jkl', 'mno', 'pqrs', 'tuv', 'wxyz']
        strs = [chars[int(d) - 2] for d in digits]
        res = []
        for s in strs:
            if not res:
                res = list(s)
            else:
                cache = []
                for item in res:
                    for letter in s:
                        cache.append(item + letter)
                res = cache
        return res

Java

class Solution {
    public List<String> letterCombinations(String digits) {
        int n;
        if ((n = digits.length()) == 0) return Collections.emptyList();
        List<String> chars = Arrays.asList("abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz");

        List<String> strs = new ArrayList<>();
        for (char c : digits.toCharArray()) {
            strs.add(chars.get(c - '0' - 2));
        }
        List<String> res = new ArrayList<>();
        for (String str : strs) {
            if (res.size() == 0) {
                for (char c : str.toCharArray()) {
                    res.add(String.valueOf(c));
                }
            } else {
                List<String> cache = new ArrayList<>();
                for (String item : res) {
                    for (char c : str.toCharArray()) {
                        cache.add(item + String.valueOf(c));
                    }
                }
                res = cache;
            }
        }
        return res;
    }
}

...