Special backtracking Leetcode of -17. A telephone number, a letter (Letter Combinations of a Phone Number)

Leetcode [17]. A telephone number, a letter (Letter Combinations of a Phone Number)

 

Subject description:

Given numbers only a  2-9 string, it returns all letter combinations indicated.

Given digital map to letters as follows (the same telephone key). Note 1 does not correspond to any alphabet.

 

 

Example:

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

 

analysis:

Title requirements of a given length comprising 2-9 N number of strings, each number corresponding to the letters, all required combinations of these letters, as shown in FIG.

We can use the Map that stores digital key 2-9, with the digital value stored in the corresponding letter (e.g., corresponding to 2 abc).

 

Recursion can solve this problem, the first index string entered = 0, according to the value argument (e.g. key = 2, value = "abc") corresponding to the key on the Map, the letters were taken, and into the next process.

 

Map storage process with the following:

 Map<String, String> map = 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");
  }};

 

 

You can see, the answer requires a List <String> as a return, then we can:

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

 

 

 

 

Next to the write function, we create a function called the dfs:

 

 

public void dfs(String digits, int step, String answer) {
        if (step == digits.length()) {
            ans.add(answer);
            return;
        }

        char c = digits.charAt(step);
        String value = map.get(c +"");
        for (int i = 0; i < value.length(); i++) {
            dfs(digits, step + 1, answer + value.charAt(i));
        }
    }

 

 

 

 

All together, and finally AC code:

 

class Solution {
    List<String> ans = new ArrayList<>();
    Map<String, String> map = 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");
          }};
    public List<String> letterCombinations(String digits) {
        if (digits.length() == 0 || digits == null)
            return ans;
        dfs(digits, 0, "");
        return ans;
    }

    public void dfs(String digits, int step, String answer) {
        if (step == digits.length()) {
            ans.add(answer);
            return;
        }

        char c = digits.charAt(step);
        String value = map.get(c + "");
        for (int i = 0; i < value.length(); i++) {
            dfs(digits, step + 1, answer + value.charAt(i));
        }
        
        
    }
}