Letter Combinations of a Phone Number

最新推荐文章于 2025-07-20 11:42:00 发布

原创最新推荐文章于 2025-07-20 11:42:00 发布 · 239 阅读

0 ·

CC 4.0 BY-SA版权

刷遍数专栏收录该内容

260 篇文章

订阅专栏

本文介绍了一种基于递归回溯的算法实现，用于生成输入数字字符串对应的电话键盘所有可能的字母组合。通过构建字符映射表并利用辅助递归函数完成组合生成。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

public class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> res = new LinkedList<>();
        if (digits == null || digits.length() == 0) {
            return res;
        }
        Map<Character, char[]> map = new HashMap<>();
        map.put('0', new char[0]);
        map.put('1', new char[0]);
        map.put('2', new char[]{'a','b','c'});
        map.put('3', new char[]{'d','e','f'});
        map.put('4', new char[]{'g','h','i'});
        map.put('5', new char[]{'j','k','l'});
        map.put('6', new char[]{'m','n','o'});
        map.put('7', new char[]{'p','q','r','s'});
        map.put('8', new char[]{'t','u','v'});
        map.put('9', new char[]{'w','x','y','z'});
        StringBuilder sb = new StringBuilder();
        helper(res, digits, map, sb, 0);
        return res;
    }
    
    private void helper(List<String> res, String digits, Map<Character, char[]> map, StringBuilder sb, int pos) {
        if (sb.length() == digits.length()) {
            res.add(new String(sb.toString()));
            return;
        }
        for (int i = pos; i < digits.length(); i++) {
            char num = digits.charAt(i);
            char[] chars = map.get(num);
            for (char c: chars) {
                sb.append(c);
                helper(res, digits, map, sb, i + 1);
                sb.deleteCharAt(sb.length() - 1);
            }
        }
    }
}