0017_Letter Combinations of a Phone Number

最新推荐文章于 2025-07-20 11:42:00 发布
最新推荐文章于 2025-07-20 11:42:00 发布
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分类专栏： LeetCode 文章标签： java
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本文链接：https://blog.youkuaiyun.com/wsy2846513/article/details/72938544

JAVA

方法一

要求数字对应的全部字母的组合，直接递归就可以了，不过效率不够高，大概排在中间的位置，要考虑优化。而且在写代码的时候，对substring，list的应用不熟，记不清关键字的拼写。。尴尬。。

public class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> result = new ArrayList<String>();
        if(digits.length() == 0){
            return result;
        }
        getString(digits,"",result);
        return result;
    }
    public void getString(String digits,String tempResult,List<String> result){
        if(digits.length() == 0){
            result.add(tempResult);
            return;
        }
        Character c = digits.charAt(0);
        switch(c){
            // case '0':
            //     getString(digits.substring(1,digits.length()),tempResult + " ",result);
            //     break;
            case '2':
                getString(digits.substring(1,digits.length()),tempResult + "a",result);
                getString(digits.substring(1,digits.length()),tempResult + "b",result);
                getString(digits.substring(1,digits.length()),tempResult + "c",result);
                break;
            case '3':
                getString(digits.substring(1,digits.length()),tempResult + "d",result);
                getString(digits.substring(1,digits.length()),tempResult + "e",result);
                getString(digits.substring(1,digits.length()),tempResult + "f",result);
                break;
            case '4':
                getString(digits.substring(1,digits.length()),tempResult + "g",result);
                getString(digits.substring(1,digits.length()),tempResult + "h",result);
                getString(digits.substring(1,digits.length()),tempResult + "i",result);
                break;
            case '5':
                getString(digits.substring(1,digits.length()),tempResult + "j",result);
                getString(digits.substring(1,digits.length()),tempResult + "k",result);
                getString(digits.substring(1,digits.length()),tempResult + "l",result);
                break;
            case '6':
                getString(digits.substring(1,digits.length()),tempResult + "m",result);
                getString(digits.substring(1,digits.length()),tempResult + "n",result);
                getString(digits.substring(1,digits.length()),tempResult + "o",result);
                break;
            case '7':
                getString(digits.substring(1,digits.length()),tempResult + "p",result);
                getString(digits.substring(1,digits.length()),tempResult + "q",result);
                getString(digits.substring(1,digits.length()),tempResult + "r",result);
                getString(digits.substring(1,digits.length()),tempResult + "s",result);
                break;
            case '8':
                getString(digits.substring(1,digits.length()),tempResult + "t",result);
                getString(digits.substring(1,digits.length()),tempResult + "u",result);
                getString(digits.substring(1,digits.length()),tempResult + "v",result);
                break;
            case '9':
                getString(digits.substring(1,digits.length()),tempResult + "w",result);
                getString(digits.substring(1,digits.length()),tempResult + "x",result);
                getString(digits.substring(1,digits.length()),tempResult + "y",result);
                getString(digits.substring(1,digits.length()),tempResult + "z",result);
                break;
        }
    }
}

方法二

考虑用栈配合循环来代替递归，觉得这样可以快一些。
使用一个字符数组来模拟栈，为了减少时间，不手工初始化数组，直接使用默认初始化的值。
然而。。用时4ms，beats 47.1%。。。。印象中把递归转换成栈会节约很多时间的呀。。

public class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> result = new ArrayList<String>();
        if(digits.length() == 0){
            return result;
        }
        char [] temp = new char[digits.length()];
        int index = 0;
        while(index >= 0){
            if(index == digits.length()){
                result.add(new String(temp));
                --index;
            }
            switch (digits.charAt(index)){
                case '2':
                    switch(temp[index]){
                        case 0: temp[index] = 'a';++index;break;
                        case 'a': temp[index] = 'b';++index;break;
                        case 'b': temp[index] = 'c';++index;break;
                        case 'c': temp[index] = 0;--index;break;
                    }
                    break;
                case '3':
                    switch(temp[index]){
                        case 0: temp[index] = 'd';++index;break;
                        case 'd': temp[index] = 'e';++index;break;
                        case 'e': temp[index] = 'f';++index;break;
                        case 'f': temp[index] = 0;--index;break;
                    }
                    break;
                case '4':
                    switch(temp[index]){
                        case 0: temp[index] = 'g';++index;break;
                        case 'g': temp[index] = 'h';++index;break;
                        case 'h': temp[index] = 'i';++index;break;
                        case 'i': temp[index] = 0;--index;break;
                    }
                    break;
                case '5':
                    switch(temp[index]){
                        case 0: temp[index] = 'j';++index;break;
                        case 'j': temp[index] = 'k';++index;break;
                        case 'k': temp[index] = 'l';++index;break;
                        case 'l': temp[index] = 0;--index;break;
                    }
                    break;
                case '6':
                    switch(temp[index]){
                        case 0: temp[index] = 'm';++index;break;
                        case 'm': temp[index] = 'n';++index;break;
                        case 'n': temp[index] = 'o';++index;break;
                        case 'o': temp[index] = 0;--index;break;
                    }
                    break;
                case '7':
                    switch(temp[index]){
                        case 0: temp[index] = 'p';++index;break;
                        case 'p': temp[index] = 'q';++index;break;
                        case 'q': temp[index] = 'r';++index;break;
                        case 'r': temp[index] = 's';++index;break;
                        case 's': temp[index] = 0;--index;break;
                    }
                    break;
                case '8':
                    switch(temp[index]){
                        case 0: temp[index] = 't';++index;break;
                        case 't': temp[index] = 'u';++index;break;
                        case 'u': temp[index] = 'v';++index;break;
                        case 'v': temp[index] = 0;--index;break;
                    }
                    break;
                case '9':
                    switch(temp[index]){
                        case 0: temp[index] = 'w';++index;break;
                        case 'w': temp[index] = 'x';++index;break;
                        case 'x': temp[index] = 'y';++index;break;
                        case 'y': temp[index] = 'z';++index;break;
                        case 'z': temp[index] = 0;--index;break;
                    }
                    break;
            }
        }
        return result;
    }
}
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