电话号码的字母组合

问题陈述

给定一个仅包含数字 2-9 的字符串，返回所有它能表示的字母组合。

给出数字到字母的映射如下（与电话按键相同）。注意 1 不对应任何字母。

示例：

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输入："23"
输出：["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

代码实现

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class Solution{
    public List<String> letterCombinations(String digits){
        List<String> list=new ArrayList<>();
        if(digits==null||digits.length()==0){
            return list;
        }
        Map<Character,String> map=new HashMap<>();
        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");
        backTrack(list, digits, map, 0, new StringBuilder());
        return list;
    }
    public void backTrack(List<String> list,String digits,Map<Character,String> map,int index,StringBuilder sb){
        // recursion teminator
        if (sb.length() == digits.length()) {
            list.add(sb.toString());
            return;
        }

        // process current logic
        String value = map.get(digits.charAt(index));
        for (int j = 0; j < value.length(); j++) {

            // drill down
            backTrack(list, digits, map, index + 1, sb.append(value.charAt(j)));

            // reverse states
            sb.deleteCharAt(sb.length() - 1);
        }
    }
}
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class Solution{
    public List<String> letterCombinations(String digits){
        if(digits==null||digits.length()==0) return new ArrayList<String>();
        String[] letter_Map={" ","*","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
        List<String> res=new ArrayList<>();
        res.add("");
        for(int i=0;i<digits.length();i++){//length用作函数调用时加括号
            String letters=letter_Map[digits.charAt(i)-'0'];//某字符减去0字符等于该字符对应的数字。
            int size=res.size();
            for(int j=0;j<size;j++){
                String temp=res.remove(0);
                for(int k=0;k<letters.length();k++){
                    res.add(temp+letters.charAt(k));
                }
            }
        }
    }
}

另解

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class Solution {
  Map<String, String> phone = new HashMap<String, String>() {{
    put("2", "abc");
    put("3", "def");
    put("4", "ghi");
    put("5", "jkl");
    put("6", "mno");
    put("7", "pqrs");
    put("8", "tuv");
    put("9", "wxyz");
  }};

  List<String> output = new ArrayList<String>();

  public void backtrack(String combination, String next_digits) {
    // if there is no more digits to check
    if (next_digits.length() == 0) {
      // the combination is done
      output.add(combination);
    }
    // if there are still digits to check
    else {
      // iterate over all letters which map 
      // the next available digit
      String digit = next_digits.substring(0, 1);//返回第一个数字
      String letters = phone.get(digit);
      for (int i = 0; i < letters.length(); i++) {//letters表示该数字在phone中对应的字符串
        String letter = phone.get(digit).substring(i, i + 1);
        // append the current letter to the combination
        // and proceed to the next digits
        backtrack(combination + letter, next_digits.substring(1));
      }
    }
  }

  public List<String> letterCombinations(String digits) {
    if (digits.length() != 0)
      backtrack("", digits);
    return output;
  }
}