17. Letter Combinations of a Phone Number

Problem

Given a string containing digits from 2-9 inclusive, return all possible letter combinations that the number could represent. Return the answer in any order.

A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.

Example 1:

Input: digits = "23"
Output: ["ad","ae","af","bd","be","bf","cd","ce","cf"]

Example 2:

Input: digits = ""
Output: []

Example 3:

Input: digits = "2"
Output: ["a","b","c"]

Constraints:

• 0 <= digits.length <= 4
• digits[i] is a digit in the range ['2', '9'].

Solution

這題要我們列出所有的可能字串，因此可以想到要用 backtracking 來解。在處理每個 digit 時，我們將 digit 對應的 letter 附加到 prefix 字串中，等處理到最後一個 digit 時，prefix 就會是其中一個對應字串。因為我們必須不斷地將 letter 附加到 prefix，再將 letter 從 prefix 中移除，所以我們不可以使用 String。如果使用 String 的話，每次對 String 附加或移除 letter 時，它都會建立一個新的 String，而這非常地耗效能。我們可以使用 StringBuilder 作為替代。

Backtracking

• Time:
• Space:

class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> results = new ArrayList<>();
        if (digits == null || digits.length() == 0) return results;

        Map<Character, char[]> map = buildDigitMap();
        dfs(digits, 0, map, new StringBuilder(), results);
        return results;
    }

    private Map<Character, char[]> buildDigitMap() {
        Map<Character, char[]> map = new HashMap<>();
        map.put('2', new char[]{'a', 'b', 'c'});
        map.put('3', new char[]{'d', 'e', 'f'});
        map.put('4', new char[]{'g', 'h', 'i'});
        map.put('5', new char[]{'j', 'k', 'l'});
        map.put('6', new char[]{'m', 'n', 'o'});
        map.put('7', new char[]{'p', 'q', 'r', 's'});
        map.put('8', new char[]{'t', 'u', 'v'});
        map.put('9', new char[]{'w', 'x', 'y', 'z'});
        return map;
    }

    private void dfs(String digits, int index, Map<Character, char[]> map, StringBuilder prefix, List<String> results) {
        if (index >= digits.length()) return;

        char c = digits.charAt(index);
        boolean isLast = index == digits.length() - 1;
        char[] letters = map.get(c);
        for (char d : letters) {
            prefix.append(d);
            if (isLast) {
                results.add(prefix.toString());
            } else {
                dfs(digits, index + 1, map, prefix, results);
            }
            prefix.deleteCharAt(prefix.length() - 1);
        }
    }
}

參考

• 17. Letter Combinations of a Phone Number, LeetCode.
• 17. Letter Combinations of a Phone Number, LeetCode Solutions.