Letter Combinations of a Phone Number

【题目描述】

 

 

 

Given a digit string, 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. 

Input:Digit string "23"

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

Note: Although the above answer is in lexicographical order, your answer could be in any order you want. 

 

 

 

 

 

【题目大意】

 

 

 

 

　　给定一个数字串，返回数字上所有字符的所有组合，数字到字符的映射如上图所示。 
　　注意： 尽管上面的结果以字符顺序排列的，你可以以任何顺序返回结果。

 

 

 

 

 

【本题答案】

 

 

 

package blog;

import java.util.LinkedList;
import java.util.List;

/**
 * @author yesr
 * @create 2018-04-19 下午11:40
 * @desc
 **/
public class Test0419 {
    public class Solution {

        private String[] map = {
                "abc",
                "def",
                "ghi",
                "jkl",
                "mno",
                "pqrs",
                "tuv",
                "wxyz",
        };
        private List<String> result;    // 存储最终结果
        private char[] chars;           // 保存去掉0，1字符的结果
        private char[] curResult;       // 存储中间结果
        private int end = 0;            // 字符数组中的第一个未使用的位置
        private int handle = 0;         // 当前处理的是第几个字符数字

        public List<String> letterCombinations(String digits) {
            result = new LinkedList<>();

            if (digits != null && digits.length() > 0) {

                chars = digits.toCharArray();

                // 对字符串进行处理，去掉0和1
                // 找第一个0或者1的位置
                while (end < digits.length() && chars[end] != '0' && chars[end] != '1') {
                    end++;
                }

                handle = end + 1;
                while (handle < chars.length) {
                    if (chars[handle] != '0' && chars[handle] != '1') {
                        chars[end] = chars[handle];
                        end++;
                    }
                    handle++;
                }

                curResult = new char[end];
                // while结束后，end为有效字符的长度
                handle = 0; // 指向第一个有效字符的位置

                letterCombinations();
            }
            return result;
        }

        private void letterCombinations() {
            if (handle >= end) {
                result.add(new String(curResult));
            } else {
                int num = chars[handle] - '2';
                for (int i = 0; i < map[num].length(); i++) {
                    curResult[handle] = map[num].charAt(i);
                    handle++;
                    letterCombinations();
                    handle--;
                }
            }
        }
    }
}