leetcode17- Letter Combinations of a Phone Number- medium

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.

算法：DFS。函数头是private void dfs(String digits, int idx, String[] map, StringBuilder crt, List<String> result) 

每次对当前这个数字可能表示的几种字母for循环去尝试。

细节：1.StringBuilder回溯的函数是sb.deleteCharAt(index)！！没有remove方法，只有delete(区间), deleteCharAt(位置)方法。 2.JAVA没有Integer.parseInt(Char)方法，可能是觉得char转int足够了就用c-'0'. 3.小心0，1的处理，不是就直接返回了，而是也要dfs下一个数字再返回。 4.出口不是看做出来的字符串的长度了，而是看现在遍历数字遍历完了没有

实现：

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

        String[] map = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        dfs(digits, 0, map, new StringBuilder(), result);
        return result;
    }

    private void dfs(String digits, int idx, String[] map, StringBuilder crt, List<String> result) {
        if (idx >= digits.length() && crt.length() > 0) {
            result.add(crt.toString());
            return;
        }

        // Integer.parseInt居然神坑，不支持转char
        int digit =digits.charAt(idx) - '0';
        if (digit == 0 || digit == 1) {
            dfs(digits, idx + 1, map, crt, result);
            return;
        }

        String chars = map[digit];
        for (int i = 0; i < chars.length(); i++) {
            crt.append(chars.charAt(i));
            dfs(digits, idx + 1, map, crt, result);
            crt.deleteCharAt(crt.length() - 1);
        }

    }
}