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"].
回溯法
说明
返回条件: 当所得到的string长度与给入的string digits长度相同时候
传入参数: 用一个数来记录string的长度, 每次递归传入长度+1和新的string
复杂度
假设有k个digit 每个digit可以代表m个字符, 时间O(m^k) 空间O(m^k)
代码
public List<String> letterCombinations(String digits) {
List<String> res = new ArrayList<String>();
if (digits == null || digits.length() == 0) {
return res;
}
String[] map= {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
StringBuilder tem = new StringBuilder();
helper(res, tem, digits, 0, map);
return res;
}
public void helper(List<String> res, StringBuilder tem, String digits, int index, String[] map) {
if (index == digits.length()) {
res.add(tem.toString());
return;
}
int m = digits.charAt(index) - '0';
for (int i = 0; i < map[m].length(); i++) {
tem.append(map[m].charAt(i));
helper(res, tem, digits, index + 1, map);
tem.deleteCharAt(tem.length() - 1);
}
}
**粗体** _斜体_ [链接](http://example.com) `代码` - 列表 > 引用
。你还可以使用@
来通知其他用户。