17. Letter Combinations of a Phone Number

Medium

好久没写过recursion感觉什么backtracking这些都忘了怎么写了，这道题拿来回顾一下吧。画一画recursion tree. (画不下，b开头的答案没画）

image.png

helper函数的定义就是当前能组成的字符是sb, 当前所有储存的答案res,当前数字和char对应的map和给予的digits下，能组成哪些新的string，把这些string加到res里面。

注意几点：

• List要拿几个元素来initialize的话，要像这样写：
  map.put(2, new ArrayList<>(Arrays.asList('a','b','c')));
• char to int的method：
  Character.getNumericValue(char c);

class Solution {
    public List letterCombinations(String digits) {
        List res = new ArrayList<>();
        if (digits == null || digits.equals("")){
            return res;
        }
        Map> map = new HashMap<>();
        map.put(2, new ArrayList<>(Arrays.asList('a','b','c')));
        map.put(3, new ArrayList<>(Arrays.asList('d','e','f')));
        map.put(4, new ArrayList<>(Arrays.asList('g','h','i')));
        map.put(5, new ArrayList<>(Arrays.asList('j','k','l')));
        map.put(6, new ArrayList<>(Arrays.asList('m','n','o')));
        map.put(7, new ArrayList<>(Arrays.asList('p','q','r','s')));
        map.put(8, new ArrayList<>(Arrays.asList('t','u','v')));
        map.put(9, new ArrayList<>(Arrays.asList('w','x','y','z')));
        StringBuilder sb = new StringBuilder();
        helper(digits, sb, map, res);
        return res;
    }
    
    private void helper(String digits, StringBuilder sb, Map> map, List res){
        if (sb.length() == digits.length()){
            res.add(sb.toString());
            return;
        }
        for (char c : map.get(Character.getNumericValue(digits.charAt(sb.length())))){
            sb.append(c);
            helper(digits,sb,map,res);
            sb.deleteCharAt(sb.length() - 1);
        }
    }
}