题解 | #电话号码的字母组合#

import java.util.*;


public class Solution {
    
    public ArrayList<String> ans = new ArrayList<>();

    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * @param num string字符串
     * @return string字符串ArrayList
     */
    public ArrayList<String> phoneNumber(String num) {
        // write code here
        HashMap<Character, ArrayList<Character>> hashMap = new HashMap<>();
        int index = 0;
        for (int i = 2; i <= 9; i++) {
            ArrayList<Character> tmpArr = new ArrayList<>();
            if (i == 7 || i == 9) {
                for (int j = 0; j < 4; j++) {
                    tmpArr.add((char) ('A' + index));
                    index++;
                }
            } else {
                for (int j = 0; j < 3; j++) {
                    tmpArr.add((char) ('A' + index));
                    index++;
                }
            }
            hashMap.put((char) ('0' + i), tmpArr);
        }
        char[] nums = num.toCharArray();
        process("", nums, 0, hashMap);
        return ans;
    }

    public void process(String previousString, char[] nums, int index, HashMap<Character, ArrayList<Character>> hashMap) {
        if (index >= nums.length) {
            ans.add(new String(previousString));
            return;
        }
        char currentChar = nums[index];
        ArrayList<Character> chrs = hashMap.get(currentChar);
        for (char chr : chrs) {
            previousString += chr;
            process(previousString, nums, index + 1, hashMap);
            previousString = previousString.substring(0, previousString.length() - 1);
        }
    }
}