Letter Combinations of a Phone Number
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"]. Note: Although the above answer is in lexicographical order, your answer could be in any order you want.
Tips:
FIFO queue, faster: 以234为例,先加入abc,然后提出a,加d,e,f存,再提出b,c,加d,e,f存,再提出ad,存adg,adh,adi,以此类推。
DFS:就是普通的dfs。。。
Code:
FIFO queue:
public class Solution {
public List<String> letterCombinations(String digits) {
List<String> ans = new ArrayList<String>();
if (digits.length() == 0) {
return ans;
}
String[] mapping = new String[] {"0", "1", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
ans.add("");
for (int i = 0; i < digits.length(); i++) {
int cur = Character.getNumericValue(digits.charAt(i));
while (ans.get(0).length() == i) {
String prev = ans.remove(0);
for (char c : mapping[cur].toCharArray()) {
ans.add(prev + c);
}
}
}
return ans;
}
}
DFS:
public class Solution {
private static final String[] KEYS = { "", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz" };
public List<String> letterCombinations(String digits) {
List<String> ret = new LinkedList<String>();
combination("", digits, 0, ret);
return ret;
}
private void combination(String prefix, String digits, int offset, List<String> ret) {
if (offset >= digits.length()) {
ret.add(prefix);
return;
}
String letters = KEYS[(digits.charAt(offset) - '0')];
for (int i = 0; i < letters.length(); i++) {
combination(prefix + letters.charAt(i), digits, offset + 1, ret);
}
}
}