17. Letter Combinations of a Phone Number

17. Letter Combinations of a Phone Number

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.

 

Example:

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

public class Solution {
    public List<String> letterCombinations(String digits) {
    //存放答案的list
        List<String> list=new ArrayList<String>();
        //建立map存放手机按键
        Map<String, String> map=new HashMap<String, 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");
        //循环判断
        for (int i = 0; i < digits.length(); i++) {
            String s=digits.substring(i,i+1);
            //如果能查到数字键盘
            //if (map.get(s)!=null) {
                //如果list不为空，即并非刚开始
                if (list.size()!=0) {
                    //记住目前list的长度
                    int size=list.size();
                    //循环list当前长度
                    for (int j = 0; j < size; j++) {
                        //将list中的每个值与接下来的对应数字键盘的可能性进行拼接
                        for (int k = 0; k <map.get(s).length(); k++) {
                            String a=list.get(j)+map.get(s).charAt(k);
                            list.add(a);
                        }
                    }
                    //删掉之前的list
                    for (int j = size-1; j >=0; j--) {
                        list.remove(j);
                    }
                }else {
                    for (int j = 0; j < map.get(s).length(); j++) {
                        String a=map.get(s).charAt(j)+"";
                        list.add(a);
                    }
                }
            //}
        }
        return list;
    }
}

 https://www.youtube.com/watch?v=uMmFXWs_ZMY

public class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> result=new ArrayList<String>();
        if(digits==null||digits.length()==0||digits.contains("0")||digits.contains("1")){
            return result;
        }
        String[] map=new String[10];
        map[0]="";
        map[1]="";
        map[2]="abc";
        map[3]="def";
        map[4]="ghi";
        map[5]="jkl";
        map[6]="mno";
        map[7]="qprs";
        map[8]="tuv";
        map[9]="wxyz";
        StringBuilder sb=new StringBuilder();
        backtrack(sb,digits,map,result,0);
        return result;
        
    }
    
    public void backtrack(StringBuilder sb,String digits,String[] map,List<String> result,int start){
        if(sb.length()==digits.length()){
            result.add(sb.toString());
            return;
        }
        for(int i=start;i<digits.length();i++){
            int tt=digits.charAt(i) - '0';
            for(int j=0;j<map[tt].length();j++){
                sb.append(map[tt].charAt(j));
                backtrack(sb,digits,map,result,i+1);
                sb.deleteCharAt(sb.length()-1);//剪枝
            }
        }
    }
}

回溯法，精华是剪枝