[LeetCode]--17. 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.

方法一：递归调用

public List<String> letterCombinations(String digits) {        List<String> result = new ArrayList<String>();        String[] map = new String[10];        map[0] = "";        map[1] = "";        map[2] = "abc";        map[3] = "def";        map[4] = "ghi";        map[5] = "jkl";        map[6] = "mno";        map[7] = "pqrs";        map[8] = "tuv";        map[9] = "wxyz";        char[] middleTemp = new char[digits.length()];        dfsGetStr(digits, 0, middleTemp, map, result);        return result;    }    private void dfsGetStr(String digits, int i, char[] middleTemp,            String[] map, List<String> result) {        if (i == digits.length()) {            result.add(new String(middleTemp));            return;        }        char strChar = digits.charAt(i);        for (int j = 0; j < map[strChar - '0'].length(); j++) {            middleTemp[i] = map[strChar - '0'].charAt(j);            dfsGetStr(digits, i + 1, middleTemp, map, result);        }    }

方法二：

public List<String> letterCombinations(String digits) {        List<String> result = new ArrayList<String>();        if (digits.isEmpty())            return result;        String[] map = new String[10];        map[0] = "";        map[1] = "";        map[2] = "abc";        map[3] = "def";        map[4] = "ghi";        map[5] = "jkl";        map[6] = "mno";        map[7] = "pqrs";        map[8] = "tuv";        map[9] = "wxyz";        int len = digits.length();        int[] number = new int[len];        int k = len - 1;        while (k >= 0) {            k = len - 1;            char[] charTemp = new char[len];            for (int i = 0; i < len; i++) {                charTemp[i] = map[digits.charAt(i) - '0'].charAt(number[i]);            }            result.add(new String(charTemp));            while (k >= 0) {                if (number[k] < (map[digits.charAt(k) - '0'].length() - 1)) {                    number[k]++;                    break;                } else {                    number[k] = 0;                    k--;                }            }        }        return result;    }