Given a string containing digits from 2-9 inclusive, 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. Note that 1 does not map to any letters.

use hashmap to store digit and characters accordingly.

at ith position of input:

　　use queue to store the combinations of the first i digits, when adding new digits, just need to get every combination in the queue and add the new digit on each one.

class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> list=new
 
 ArrayList<>();
        if(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");
        Queue<String> queue=new LinkedList<>();
        queue.offer("");
        for(char c:digits.toCharArray()){
            int num=queue.size();
            for(int i=0;i<num;i++){
                String str=queue.poll();
                for(char x:map.get(c).toCharArray()){
                    queue.offer(str+x);
                }
            }
        }
        while(!queue.isEmpty()){
            list.add(queue.poll());
        }
        return list;
    }
}

不要老是想著hashmap！

如果index可以用數組下標表示就避免用hashmap，數組的維護比hashmap要容易快捷的多

class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> list=new ArrayList<>();
        if(digits.length()==0) return list;
        String[] record=new String[]{null, null, "abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
        Queue<String> queue=new LinkedList<>();
        queue.offer("");
        for(char c:digits.toCharArray()){
            int num=queue.size();
            for(int i=0;i<num;i++){
                String str=queue.poll();
                for(char x:record[c-‘0‘].toCharArray()){
                    queue.offer(str+x);
                }
            }
        }
        while(!queue.isEmpty()){
            list.add(queue.poll());
        }
        return list;
    }
}

backtracking:

class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> list=new ArrayList<>();
        if(digits.length()==0) return list;
        String[] record=new String[]{null, null, "abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
        getCombination(digits, record, "", 0, list);
        return list;
    }
    public void getCombination(String digits, String[] record, String str, int index, List<String> list){
        if(index>=digits.length()){
            list.add(str);
            return;
        }
        for(int i=0;i<record[digits.charAt(index)-‘0‘].length();i++){
            getCombination(digits, record, str+record[digits.charAt(index)-‘0‘].charAt(i), index+1, list);
        }
    }
}

leetcode 17-Letter Combinations of a Phone Number(medium)