Letter Combinations of a Phone Number

Letter Combinations of a Phone Number 

问题：

iven 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.

思路：

　　dfs+回溯

我的代码：

public class Solution {

    public List<String> letterCombinations(String digits) {

        List<String> list = new ArrayList<String>();

        if(digits == null || digits.length() == 0)  return list;

        helper("", digits, list);

        return list;

    }

    public void helper(String s, String digits, List<String> list)

    {

        if(digits.length() == 0)

        {

            list.add(s);

            return;

        }

        char digit = digits.charAt(0);

        String letters = getLetters(digit);

        if(letters.equals(""))

        {

            helper(s, digits.substring(1), list);

        }

        for(int i = 0; i <letters.length(); i++)

        {

            helper(s+letters.charAt(i), digits.substring(1), list);

        }

    }

    public String getLetters(char c)

    {

        switch(c)

        {

            case '2': return "abc";

            case '3': return "dfe";

            case '4': return "ghi";

            case '5': return "jkl";

            case '6': return "mno";

            case '7': return "pqrs";

            case '8': return "tuv";

            case '9': return "wxyz";

            default : return "";

        }

    }

}

View Code

他人代码：

public class Solution {

    public ArrayList<String> letterCombinations(String digits) {

        ArrayList<String> result = new ArrayList<String>();



        if (digits == null) {

            return result;

        }



        Map<Character, char[]> map = new HashMap<Character, char[]>();

        map.put('0', new char[] {});

        map.put('1', new char[] {});

        map.put('2', new char[] { 'a', 'b', 'c' });

        map.put('3', new char[] { 'd', 'e', 'f' });

        map.put('4', new char[] { 'g', 'h', 'i' });

        map.put('5', new char[] { 'j', 'k', 'l' });

        map.put('6', new char[] { 'm', 'n', 'o' });

        map.put('7', new char[] { 'p', 'q', 'r', 's' });

        map.put('8', new char[] { 't', 'u', 'v'});

        map.put('9', new char[] { 'w', 'x', 'y', 'z' });



        StringBuilder sb = new StringBuilder();

        helper(map, digits, sb, result);



        return result;

    }



    private void helper(Map<Character, char[]> map, String digits, 

        StringBuilder sb, ArrayList<String> result) {

        if (sb.length() == digits.length()) {

            result.add(sb.toString());

            return;

        }



        for (char c : map.get(digits.charAt(sb.length()))) {

            sb.append(c);

            helper(map, digits, sb, result);

            sb.deleteCharAt(sb.length() - 1);

        }

    }

}

View Code

学习之处：

• string 自带回溯属性 这方面减少不少的问题 不断的新建s 消耗的空间还是挺多的。
• 他人代码里面用的是StringBuffer相对而言，节约了空间。