【LeetCode】#17电话号码的组合(Letter Combinations of a Phone Number)

【LeetCode】#17电话号码的组合(Letter Combinations of a Phone Number)

题目描述

给定一个仅包含数字 2-9 的字符串，返回所有它能表示的字母组合。
给出数字到字母的映射如下（与电话按键相同）。注意 1 不对应任何字母。

示例

输入：“23”
输出：[“ad”, “ae”, “af”, “bd”, “be”, “bf”, “cd”, “ce”, “cf”].

Description

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”].

解法

class Solution {
    public List<String> letterCombinations(String digits) {

        List<String>list=new ArrayList<>();
        String []s=new String[digits.length()];
        int M=digits.length();
        if(s.length==0){
            return list;
        }
        for(int i=0;i<digits.length();i++){
            switch (digits.charAt(i)){
                case '2':s[i]="abc";break;
                case '3':s[i]="def";break;
                case '4':s[i]="ghi";break;
                case '5':s[i]="jkl";break;
                case '6':s[i]="mno";break;
                case '7':s[i]="pqrs";break;
                case '8':s[i]="tuv";break;
                case '9':s[i]="wxyz";break;
            }
        }
        list=getStringWithFor(s,0,list,"");
        return list;
    }

    private static List<String> getStringWithFor(String []s,int i,List<String> list,String stemp) {

        if(i<s.length-1){
            for(int j=0;j<s[i].length();j++){
                list=getStringWithFor(s,i+1,list,stemp+s[i].charAt(j));
            }
            i++;
        } else {
            for(int j=0;j<s[i].length();j++){
                list.add(stemp+s[i].charAt(j));
            }
        }
        return list;
    }
}