🌕 17. 电话号码的字母组合

难度: 🌕

问题描述

---

解法

class Solution {
    List<String> res = new ArrayList<>();
    String[] array = new String[] {"#", "#", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
    public List<String> letterCombinations(String digits) {
        // 思路:
        // 将电话号码信息存入数组
        // 回溯
        int len = digits.length();
        // 特殊情况特判
        if(len == 0) {
            return res;
        }
        LinkedList<Character> path = new LinkedList<>();
        mySol(digits, len, 0, path);
        return res;
    }

    private void mySol(String digits, int len, int index, LinkedList<Character> path) {
        // 递归终止条件
        if(index == len) {
            res.add(getStr(path));
            return;
        }
        // index < len
        String str = array[digits.charAt(index) - '0'];
        for(int i = 0; i < str.length(); i ++) {
            path.addLast(str.charAt(i));
            mySol(digits, len, index + 1, path);
            path.removeLast();
        }
    }

    private String getStr(LinkedList<Character> path) {
        StringBuilder sb = new StringBuilder();
        for(int i = 0; i < path.size(); i ++) {
            sb.append(path.get(i));
        }
        return sb.toString();
    }
}

---

输出