17.Letter Combinations of a Phone Number

给定一个值为 2~ 9 的字符串,每个数字如下所示,输出所有可能的组合结果。


Input: "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

 

思路:
一开始没读懂题意,以为是两两结合,其实是所有数字一起的组合。比如“234”,输出模板为“adg"。鉴于可以出现一个数字、多个数字,所以遍历的时候,将之前容器 res存的结果取出,在每个结果的末尾加上当前数字代表的字母,然后再放入新的容器 tmp ,每次循环完 res = tmp; tmp.clear( )。由于第一次循环的时候,res还是空的,防止不能进循环,所以初始res = {""}.

vector<string> letterCombinations(string digits) {
    if (digits.size() == 0) return {};
    string a[] = { "","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz" };
    vector<string> res{""}; //防止首次不能正常循环
    int n = digits.size();
    for (int i = 0; i < n; i++) {
        vector<string> tmp;
        string s = a[digits[i] - '0'];
        int len_s = s.size(), len_res = res.size();
        for (auto res_item : res) {
            for (auto s_item : s) {
                tmp.push_back(res_item + s_item);
            }
        }
        res = tmp;
        tmp.clear();
    }
    return res;
}

 

经验:对于长度不等的字符串,用string的数组存储,比用容器或二维数组好些。 

 

Java 版:

class Solution {
    public List<String> letterCombinations(String digits) {
        if(digits.length() == 0) return new ArrayList<String>();
        List<String> res = new ArrayList<>();
        List<String> tmpResult = new ArrayList<>();
        res.add("");
        char[][] template = {{'a','b','c'},
            {'d','e','f'},
            {'g','h','i'},
            {'j','k','l'},
            {'m','n','o'},
            {'p','q','r','s'},
            {'t','u','v'},
            {'w','x','y','z'}
        };
        for(int i = 0; i < digits.length(); i++){
            int n = digits.charAt(i) - '0';
            for(String s : res){
                for(char c : template[n-2])
                    tmpResult.add(s + c);
            }
            res.clear();
            res.addAll(tmpResult);
            tmpResult.clear();
        }
        return res;
    }
}

 

posted @ 2020-05-18 18:10  星海寻梦233  阅读(110)  评论(0编辑  收藏  举报