Backtracking - 17. Letter combinations of phone numbers

1 Title Description

Given a number only 2-9 String , Returns all the letter combinations it can represent . You can press In any order return .

The mapping of numbers to letters is given as follows （ Same as phone key ）. Be careful 1 Does not correspond to any letter .

source ： Power button （LeetCode）
link ：https://leetcode.cn/problems/letter-combinations-of-a-phone-number

2 Title Example

Example 1：
Input ：digits = “23”
Output ：[“ad”,“ae”,“af”,“bd”,“be”,“bf”,“cd”,“ce”,“cf”]

Example 2：
Input ：digits = “”
Output ：[]

Example 3：
Input ：digits = “2”
Output ：[“a”,“b”,“c”]

3 Topic tips

0 <= digits.length <= 4
digits[i] Is the scope of [‘2’, ‘9’] A number of .

4 Ideas

First, use a hash table to store all possible letters corresponding to each number , And then do the backtracking operation .
Maintaining a string during backtracking , Represents an existing alphabetic arrangement ( If you don't traverse all the numbers of the phone number , The existing alphabetic arrangement is incomplete ). The string is initially empty . Take one digit of the phone number at a time , Get all possible letters corresponding to the number from the hash table , And insert one of the letters after the existing alphabet , Then continue to process the last digit of the phone number , Until all the numbers in the phone number are processed , That is to get a complete alphabetic arrangement . And then go back , Traverse the rest of the alphabet .
Backtracking algorithm is used to find all feasible solutions , If a solution is found to be infeasible , It will abandon the infeasible solution . In this question , Because every letter corresponding to every number can enter the alphabet , So there is no infeasible solution , Just enumerate all the solutions directly .
Complexity analysis

· Time complexity :O(3^m× 4^n), among m Is the input corresponding to 3 The number of letters ( Including digital 2、3、4、5、6、8),n Is the input corresponding to 4 The number of letters （ Including digital 7、9) , m＋n Is the total number of input numbers . When the input contains m A correspondence 3 Two letter numbers and n A correspondence 4 It's a letter number , Different letter combinations — share 3^m×4n Kind of , Need to traverse every — Alphabetic combination .
Spatial complexity :O(m + n), among m Is the input corresponding to 3 The number of letters ,n Is the input corresponding to 4 The number of letters ,m＋n Is the total number of input numbers . In addition to the return value , The spatial complexity mainly depends on the hash table and the number of recursive calls in the backtracking process , The size of the hash table is independent of the input , You can think of it as a constant , The maximum number of recursive calls is m ＋n.

5 My answer

class Solution {
    
    public List<String> letterCombinations(String digits) {
    
        List<String> combinations = new ArrayList<String>();
        if (digits.length() == 0) {
    
            return combinations;
        }
        Map<Character, String> phoneMap = new HashMap<Character, 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");
        }};
        backtrack(combinations, phoneMap, digits, 0, new StringBuffer());
        return combinations;
    }

    public void backtrack(List<String> combinations, Map<Character, String> phoneMap, String digits, int index, StringBuffer combination) {
    
        if (index == digits.length()) {
    
            combinations.add(combination.toString());
        } else {
    
            char digit = digits.charAt(index);
            String letters = phoneMap.get(digit);
            int lettersCount = letters.length();
            for (int i = 0; i < lettersCount; i++) {
    
                combination.append(letters.charAt(i));
                backtrack(combinations, phoneMap, digits, index + 1, combination);
                combination.deleteCharAt(index);
            }
        }
    }
}

Give a funny solution ：

class Solution {
    

    public List<String> letterCombinations(String digits) {
    
        List<String> list = new ArrayList<>();
        StringBuilder sb = new StringBuilder();
        if(digits.length() == 0){
    
            return list;
        }

        backtracking(digits, 0, list, sb);

        return list;
    }

    public void backtracking(String digits, int index, List<String> list, StringBuilder sb){
    
        if(sb.length() == digits.length()){
    
            list.add(sb.toString());
            return;
        }

        for(int i = index; i<digits.length(); i++){
    
            char c = digits.charAt(i); 
            if(c == '2'){
    
                sb.append('a');
                backtracking(digits, i+1, list, sb);
                sb.deleteCharAt(sb.length()-1);
                sb.append('b');
                backtracking(digits, i+1, list, sb);
                sb.deleteCharAt(sb.length()-1);
                sb.append('c');
                backtracking(digits, i+1, list, sb);
                sb.deleteCharAt(sb.length()-1);
            }else if(c == '3'){
    
                sb.append('d');
                backtracking(digits, i+1, list, sb);
                sb.deleteCharAt(sb.length()-1);
                sb.append('e');
                backtracking(digits, i+1, list, sb);
                sb.deleteCharAt(sb.length()-1);
                sb.append('f');
                backtracking(digits, i+1, list, sb);
                sb.deleteCharAt(sb.length()-1);
            }else if(c == '4'){
    
                sb.append('g');
                backtracking(digits, i+1, list, sb);
                sb.deleteCharAt(sb.length()-1);
                sb.append('h');
                backtracking(digits, i+1, list, sb);
                sb.deleteCharAt(sb.length()-1);
                sb.append('i');
                backtracking(digits, i+1, list, sb);
                sb.deleteCharAt(sb.length()-1);
            }else if(c == '5'){
    
                sb.append('j');
                backtracking(digits, i+1, list, sb);
                sb.deleteCharAt(sb.length()-1);
                sb.append('k');
                backtracking(digits, i+1, list, sb);
                sb.deleteCharAt(sb.length()-1);
                sb.append('l');
                backtracking(digits, i+1, list, sb);
                sb.deleteCharAt(sb.length()-1);
            }else if(c == '6'){
    
                sb.append('m');
                backtracking(digits, i+1, list, sb);
                sb.deleteCharAt(sb.length()-1);
                sb.append('n');
                backtracking(digits, i+1, list, sb);
                sb.deleteCharAt(sb.length()-1);
                sb.append('o');
                backtracking(digits, i+1, list, sb);
                sb.deleteCharAt(sb.length()-1);
            }else if(c == '7'){
    
                sb.append('p');
                backtracking(digits, i+1, list, sb);
                sb.deleteCharAt(sb.length()-1);
                sb.append('q');
                backtracking(digits, i+1, list, sb);
                sb.deleteCharAt(sb.length()-1);
                sb.append('r');
                backtracking(digits, i+1, list, sb);
                sb.deleteCharAt(sb.length()-1);
                sb.append('s');
                backtracking(digits, i+1, list, sb);
                sb.deleteCharAt(sb.length()-1);
            }else if(c == '8'){
    
                sb.append('t');
                backtracking(digits, i+1, list, sb);
                sb.deleteCharAt(sb.length()-1);
                sb.append('u');
                backtracking(digits, i+1, list, sb);
                sb.deleteCharAt(sb.length()-1);
                sb.append('v');
                backtracking(digits, i+1, list, sb);
                sb.deleteCharAt(sb.length()-1);
            }else if(c == '9'){
    
                sb.append('w');
                backtracking(digits, i+1, list, sb);
                sb.deleteCharAt(sb.length()-1);
                sb.append('x');
                backtracking(digits, i+1, list, sb);
                sb.deleteCharAt(sb.length()-1);
                sb.append('y');
                backtracking(digits, i+1, list, sb);
                sb.deleteCharAt(sb.length()-1);
                sb.append('z');
                backtracking(digits, i+1, list, sb);
                sb.deleteCharAt(sb.length()-1);
            }
        }
    }

}