leetcode17+stringbuilder与递归

import java.util.ArrayList;
import java.util.List;

public class LetterCombinations_17 {
    private static final String[] KEYS = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};

    List solution(String digits){

        List combinations = new ArrayList<>();

        if ( digits==null||digits.length()==0){
            return combinations;
        }
        doCombination(new StringBuilder(),combinations,digits);


        return combinations;
    }

    private void doCombination(StringBuilder stringBuilder, List combinations, String digits) {
        if (stringBuilder.length()==digits.length()){
            combinations.add(stringBuilder.toString());
            return;
        }
        //深度优先遍历整个keys的对应index的字符串所有数字
        int curDigits = digits.charAt(stringBuilder.length())-'0';//找的是stringbuilder的长度所在的索引，在给定数字字符串位置
        String letters = KEYS[curDigits];
        for (char m : letters.toCharArray()){
            stringBuilder.append(m);//添加
            doCombination(stringBuilder,combinations,digits);
            stringBuilder.deleteCharAt(stringBuilder.length()-1);
            //删除新添加的尾部位置，23就是ad,ae,af，依次把def算完后再删掉，
            // 同理234就是3层递归，就是ad循环把adg，adh，adi的ghi删除
        }
    }

    public static void main(String[] args) {
        new LetterCombinations_17().solution("23");
        System.out.println("dd");
    }

}