官方链接
https://leetcode-cn.com/problems/letter-combinations-of-a-phone-number/
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。
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:
输入:"23"
输出:["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
解法一
backtracking
class Solution {
public List<String> letterCombinations(String digits) {
if (digits == null || digits.length() == 0) return new ArrayList<>();
Map<Character, String> map = new HashMap<>();
map.put('2', "abc");
map.put('3', "def");
map.put('4', "ghi");
map.put('5', "jkl");
map.put('6', "mno");
map.put('7', "pqrs");
map.put('8', "tuv");
map.put('9', "wxyz");
List<String> result = new ArrayList<>();
_generate(result, map, digits, 0, "");
return result;
}
public void _generate(List<String> result, Map<Character, String> map, String digits, int index, String s) {
// terminator
if (index == digits .length()) {
result.add(s);
return;
}
// process
String letters = map.get(digits.charAt(index));
for (int i = 0; i < letters.length(); i++) {
// drill down
_generate(result, map, digits, index + 1, s + letters.charAt(i));
}
// reverse states
}
}
class Solution:
def letterCombinations(self, digits: str) -> List[str]:
if not digits:
return []
self.result = []
self.d = {
'2': 'abc',
'3': 'def',
'4': 'ghi',
'5': 'jkl',
'6': 'mno',
'7': 'pqrs',
'8': 'tuv',
'9': 'wxyz'
}
self.dfs(digits, [], 0)
return self.result
def dfs(self, digits, temp, index):
if len(temp) == len(digits):
self.result.append("".join(x for x in temp))
return
for c in self.d[digits[index]]:
temp.append(c)
self.dfs(digits, temp, index+1)
temp.pop()