[LeetCode] 17. 电话号码的字母组合
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题目描述:https://leetcode-cn.com/problems/letter-combinations-of-a-phone-number/
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。
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"].
思路:
思路1:
递归
递归过程中记录组合.
思路2:
迭代,类似BFS
,每次更新一个字母
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代码:
思路1:
python
class Solution:
def letterCombinations(self, digits: str) -> List[str]:
lookup = {
"2":"abc",
"3":"def",
"4":"ghi",
"5":"jkl",
"6":"mno",
"7":"pqrs",
"8":"tuv",
"9":"wxyz"
}
if not digits:
return []
n = len(digits)
res = []
def helper(i,tmp):
if i == n:
res.append(tmp)
return
for alp in lookup[digits[i]]:
helper(i+1,tmp+alp)
helper(0,"")
return res
java
class Solution {
public List<String> letterCombinations(String digits) {
if (digits == null || digits.length() == 0) {
return new ArrayList();
}
Map<Character, String> map = new HashMap<Character, String>();
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> res = new LinkedList<String>();
helper("", digits, 0, res, map);
return res;
}
private void helper(String s, String digits, int i, List<String> res, Map<Character, String> map) {
if (i == digits.length()) {
res.add(s);
return;
}
String letters = map.get(digits.charAt(i));
for (int j = 0; j < letters.length(); j++){
helper(s+letters.charAt(j),digits,i+1,res,map);
}
}
}
思路2:
python
class Solution:
def letterCombinations(self, digits: str) -> List[str]:
lookup = {
"2":"abc",
"3":"def",
"4":"ghi",
"5":"jkl",
"6":"mno",
"7":"pqrs",
"8":"tuv",
"9":"wxyz"
}
if not digits:
return []
res = [""]
for num in digits:
next_res = []
for alp in lookup[num]:
for tmp in res:
next_res.append(tmp + alp)
res = next_res
return res
c++
vector<string> letterCombinations(string digits) {
vector<string> res;
string charmap[10] = {"0", "1", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
res.push_back("");
for (int i = 0; i < digits.size(); i++)
{
vector<string> tempres;
string chars = charmap[digits[i] - '0'];
for (int c = 0; c < chars.size();c++)
for (int j = 0; j < res.size();j++)
tempres.push_back(res[j]+chars[c]);
res = tempres;
}
return res;
}
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