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| class Solution {
Map<String, String> phone = new HashMap<String, String>() {{
put("2", "abc");
put("3", "def");
put("4", "ghi");
put("5", "jkl");
put("6", "mno");
put("7", "pqrs");
put("8", "tuv");
put("9", "wxyz");
}};
List<String> output = new ArrayList<String>();
public void backtrack(String combination, String next_digits) {
// if there is no more digits to check
if (next_digits.length() == 0) {
// the combination is done
output.add(combination);
}
// if there are still digits to check
else {
// iterate over all letters which map
// the next available digit
String digit = next_digits.substring(0, 1);
String letters = phone.get(digit);
for (int i = 0; i < letters.length(); i++) {
String letter = phone.get(digit).substring(i, i + 1);
// append the current letter to the combination
// and proceed to the next digits
// letter往后一直移动是因为从a到b,b到c， c到d。。。这样
backtrack(combination + letter, next_digits.substring(1));
// backtrack(letter + combination, next_digit.substring(1));
//Output ["da","ea","fa","db","eb","fb","dc","ec","fc"]
}
}
}
public List<String> letterCombinations(String digits) {
if (digits.length() != 0)
backtrack("", digits);
return output;
}
}
|