原题
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"].
说明:
尽管上面的答案是按字典序排列的,但是你可以任意选择答案输出的顺序。
解法
思想
dfs就嗯搜 →_→
代码
class Solution {
List<String> ans = new ArrayList<>();
char[] chars;
public List<String> letterCombinations(String digits) {
chars = digits.toCharArray();
if(chars.length == 0) return new ArrayList<String>();
List<Character> list = new ArrayList<>();
dfs(list,0);
return ans;
}
public void dfs(List<Character> list,int index){
if(index == chars.length) {
StringBuilder sb = new StringBuilder();
for(Character c:list){
sb.append(c);
}
ans.add(sb.toString());
return;
}
for(char c:numToChars(chars[index])){
List<Character> copy = new ArrayList<>(list);
copy.add(c);
dfs(copy,index+1);
}
}
public char[] numToChars(char num){
switch(num){
case '2':
return new char[]{'a','b','c'};
case '3':
return new char[]{'d','e','f'};
case '4':
return new char[]{'g','h','i'};
case '5':
return new char[]{'j','k','l'};
case '6':
return new char[]{'m','n','o'};
case '7':
return new char[]{'p','q','r','s'};
case '8':
return new char[]{'t','u','v'};
case '9':
return new char[]{'w','x','y','z'};
}
return new char[]{};
}
}
原创文章,作者:彭晨涛,如若转载,请注明出处:https://www.codetool.top/article/leetcode17-%e7%94%b5%e8%af%9d%e5%8f%b7%e7%a0%81%e7%9a%84%e5%ad%97%e6%af%8d%e7%bb%84%e5%90%88/