#leetcode#Letter Combinations of a Phone Number-优快云博客

本文链接：https://blog.youkuaiyun.com/ChiBaoNeLiuLiuNi/article/details/46754841

本文介绍了如何通过迭代和递归方法将数字字符串转换为可能的字母组合，详细阐述了电话按键对应字母的映射规则，并提供了代码实现。

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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"].

经典题目， iterative和recursive都要会写

iterative的是之前看的code ganker大神的解法，其实自己想也想得出来吧，脑子太懒了

http://blog.youkuaiyun.com/linhuanmars/article/details/19743197

recursive的自己写了一下，思路对了，没能bug free

public class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> res = new ArrayList<String>();
        if(digits == null || digits.length() == 0)
            return res;
            
        Map<Character, String> map = new HashMap<Character, String>();
        map.put('2', "abc");
        map.put('3', "def");
        map.put('4', "ghi");
        map.put('5', "jkl");
        map.put('6', "mno");
        map.put('7', "pqrs");
        map.put('8', "tuv");
        map.put('9', "wxyz");
        
        StringBuilder sb = new StringBuilder();
        
        helper(res, digits, map, sb, 0);
        
        return res;
    }
    
    private void helper(List<String> res, String digits, Map<Character, String> map, StringBuilder sb, int index){
        
        if(sb.length() == digits.length()){
            res.add(sb.toString());
            return;
        }
        
        String letters = map.get(digits.charAt(index));
        
        for(int i = 0; i < letters.length(); i++){
            sb.append(letters.charAt(i));
            helper(res, digits, map, sb, index + 1);
            sb.deleteCharAt(sb.length() - 1);
        }
    }
}