LeetCode-17 - Letter combination of phone number - Backtracking algorithm

LeetCode-17-The alphabet of the phone number

Problem description: Given a string containing only the numbers 2-9, return all the letter combinations it can represent. Answers can be returned in any order.
The mapping of numbers to letters is given below (same as phone keys). Note that 1 does not correspond to any letters.

Question link: LeetCode-17-The letter combination of the phone number

Problem-solving ideas:

1. You need to write the mapping relationship between numbers and letters yourself
2. It is an ordinary backtracking algorithm, just follow the three steps of backtracking

What's very strange is that if I write res and sb as all variables, then the results of the previous sample will appear in the results of the later sample when I submit it. I have no choice but to change it.

Code:

class Solution {
    
    
    /**
     * 1.将数字映射为对应的字母
     * 2.开始组合
     *      树的深度由输入字符串的长度控制
     *      树的宽度由digits[0]的长度控制
     */

    List<String> res = new ArrayList<>();// 结果集
    StringBuilder sb = new StringBuilder();// 相当于path，收获单个结果
    public List<String> letterCombinations(String digits) {
    
    
        if (digits.length()==0 || digits== null){
    
    
            return res;
        }
        // 自己需要将数字映射为对应的字母
        String[] numString = new String[]{
    
    "","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
        int k = digits.length();// 树的深度
        backTracking(numString, k, digits, 0);
        return res;
    }


    /**
     * @param numString 字母和数字的映射
     * @param k 树的深度，也就是digits.length()
     * @param index 表示遍历到哪个数字
     * @param digits
     */
    private void backTracking(String[] numString, int k, String digits,int index){
    
    
        if (sb.length() == k){
    
    
            res.add(sb.toString());
            return;
        }
        String singleStr = numString[digits.charAt(index)-'0'];// abc
        for (int i = 0; i < singleStr.length(); i++) {
    
    // 不需要 startIndex帮忙做去重的操作
            sb.append(singleStr.charAt(i));
            backTracking(numString, k, digits, index+1);
            sb.deleteCharAt(sb.length()-1);// 弹出去
        }
    }
}     * @param numString 字母和数字的映射
     * @param k 树的深度，也就是digits.length()
     * @param index 表示遍历到哪个数字
     * @param digits
     */
    private static void backTracking(String[] numString, int k, String digits,List<String> res, StringBuilder sb, int index){
    
    
        if (sb.length() == k){
    
    
            System.out.println(sb.toString());
            res.add(sb.toString());
            return;
        }
        String singleStr = numString[digits.charAt(index)-'0'];// abc
        for (int i = 0; i < singleStr.length(); i++) {
    
    // 不需要 startIndex帮忙做去重的操作
            backTracking(numString, k, digits, res, sb.append(singleStr.charAt(i)), index+1);
            sb.deleteCharAt(sb.length()-1);
        }
    }
}