Letter Combinations of a Phone Number

https://oj.leetcode.com/problems/letter-combinations-of-a-phone-number/

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.

解题思路：

针对输入的字符串digits，生成一个二维数组，横列是digits的每个字符，数列就是每个字符可能代表的字母。这里要注意，7和9可能代表4个字母，其他数字可能代表3个字母。

下面的问题就变成，每个字符选出一个字母，可能有多少种情况。比较典型的可以用dfs求解的问题。

public class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> returnList = new ArrayList<String>();
        if(digits.length() == 0){
　　　　　　　　return returnList;
        }
        
        char[][] chars = new char[digits.length()][4];
        for(int i = 0; i < digits.length(); i++){
            if(digits.charAt(i) >= '2' && digits.charAt(i) <= '6'){
                for(int j = 0; j < 3; j++){
                    // if(j == 3){
                    //     chars[i][j] = '1';
                    // }
                    chars[i][j] = (char)('a' + (digits.charAt(i) - '0' - 2) * 3 + j);
                }
            }
            if(digits.charAt(i) >= '7' && digits.charAt(i) <= '9'){
                if(digits.charAt(i) == '7'){
                    chars[i] = new char[]{'p','q','r','s'};
                }
                if(digits.charAt(i) == '8'){
                    chars[i] = new char[]{'t','u','v'};
                }
                if(digits.charAt(i) == '9'){
                    chars[i] = new char[]{'w','x','y','z'};
                }
            }
        }
        
        StringBuffer bf = new StringBuffer();
        dfs(returnList, chars, 0, bf);
        return returnList;
    }
    
    public static void dfs(List<String> returnList, char[][] chars, int start, StringBuffer bf){
        if(start == chars.length){
            returnList.add(bf.toString());
            return;
        }
        for(int i = 0; i < chars[start].length; i++){
            if(chars[start][i] == '\0'){
                return;
            }
            bf.append(chars[start][i]);
            dfs(returnList, chars, start + 1, bf);
            bf.deleteCharAt(start);
        }
    }
}

这里的dfs递归实际上只用到一个变量start，所以也可以将其他变量声明为成员变量。

public class Solution {
    List<String> returnList;
    char[][] chars;
    StringBuffer bf = new StringBuffer();

    public List<String> letterCombinations(String digits) {
        returnList = new ArrayList<String>();
        if(digits.length() == 0){
　　　　　　　　return returnList;
        }
        
        chars = new char[digits.length()][4];
        for(int i = 0; i < digits.length(); i++){
            if(digits.charAt(i) >= '2' && digits.charAt(i) <= '6'){
                for(int j = 0; j < 3; j++){
                    // if(j == 3){
                    //     chars[i][j] = '1';
                    // }
                    chars[i][j] = (char)('a' + (digits.charAt(i) - '0' - 2) * 3 + j);
                }
            }
            if(digits.charAt(i) >= '7' && digits.charAt(i) <= '9'){
                if(digits.charAt(i) == '7'){
                    chars[i] = new char[]{'p','q','r','s'};
                }
                if(digits.charAt(i) == '8'){
                    chars[i] = new char[]{'t','u','v'};
                }
                if(digits.charAt(i) == '9'){
                    chars[i] = new char[]{'w','x','y','z'};
                }
            }
        }
        
        dfs(0);
        return returnList;
    }
    
    public void dfs(int start){
        if(start == chars.length){
            returnList.add(bf.toString());
            return;
        }
        for(int i = 0; i < chars[start].length; i++){
            if(chars[start][i] == '\0'){
                return;
            }
            bf.append(chars[start][i]);
            dfs(start + 1);
            bf.deleteCharAt(start);
        }
    }
}

 当然，这道题也可以用bfs的方法，代码如下。

public class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> returnList = new ArrayList<String>();
        if(digits.length() == 0){
　　　　　　　　return returnList;
        }
        
        //对于2-6的数字，第四个字母为初始化，所以为'\u0000'
        char[][] chars = new char[digits.length()][4];
        for(int i = 0; i < digits.length(); i++){
            if(digits.charAt(i) >= '2' && digits.charAt(i) <= '6'){
                for(int j = 0; j < 3; j++){
                    // if(j == 3){
                    //     chars[i][j] = '1';
                    // }
                    chars[i][j] = (char)('a' + (digits.charAt(i) - '0' - 2) * 3 + j);
                }
            }
            if(digits.charAt(i) >= '7' && digits.charAt(i) <= '9'){
                if(digits.charAt(i) == '7'){
                    chars[i] = new char[]{'p','q','r','s'};
                }
                if(digits.charAt(i) == '8'){
                    chars[i] = new char[]{'t','u','v'};
                }
                if(digits.charAt(i) == '9'){
                    chars[i] = new char[]{'w','x','y','z'};
                }
            }
        }
        
        //bfs方法
        returnList.add("");
        //第一层循环是digits所含有的数字数量
        for(int i = 0; i < chars.length; i++){
            List<String> tempReturnList = new ArrayList<String>();
            //第二层循环是每个数字代表的字母数量
            for(int j = 0; j < chars[i].length; j++){
                //第三层循环是returnList里已经组成的字符串数量
                for(int k = 0; k < returnList.size(); k++){
                    if(chars[i][j] == '\0'){
                        continue;
                    }
                    tempReturnList.add(returnList.get(k) + chars[i][j]);
                }
            }
            returnList = tempReturnList;
        }
        return returnList;
    }
}

 //20181202

前面编写电话号码的简单点

public class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> returnList = new ArrayList<String>();
        if(digits.length() == 0){
            return returnList;
        }
        
        String[] telephone = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        String[] numbers = new String[digits.length()];
        for (int i = 0; i < digits.length(); i++) {
            numbers[i] = telephone[digits.charAt(i) - '0'];
        }
        
        StringBuffer bf = new StringBuffer();
        dfs(returnList, numbers, 0, bf);
        return returnList;
    }
    
    public static void dfs(List<String> returnList, String[] numbers, int start, StringBuffer bf){
        if(start == numbers.length){
            returnList.add(bf.toString());
            return;
        }
        for(int i = 0; i < numbers[start].length(); i++){
            bf.append(numbers[start].charAt(i));
            dfs(returnList, numbers, start + 1, bf);
            bf.deleteCharAt(start);
        }
    }
}