leetcode-17-电话号码的字母组合（中等)

17. 电话号码的字母组合（中等）

给定一个仅包含数字 2-9 的字符串，返回所有它能表示的字母组合。答案可以按 任意顺序 返回。

给出数字到字母的映射如下（与电话按键相同）。注意 1 不对应任何字母。

示例 1：

输入：digits = "23"
输出：["ad","ae","af","bd","be","bf","cd","ce","cf"]

示例 2：

输入：digits = ""
输出：[]

示例 3：

输入：digits = "2"
输出：["a","b","c"]

思路：

1. 递归：每次增加所digits[i]所对应的数字的可选字符中的一个加入temp中，通过递归，可得到最终结果。、
2. 迭代法：依次根据digits字符串的数字顺序从左到右遍历的顺序往⾥⾯添加字母，每一次都追加在原有res的后面，因为res会改变所以每次设立一个空string数组temp，然后temp根据result中原有的结果向后面继续添加拼接原+新的字符串，然后res = temp进行复制

1. int('6')的结果为字符6所对应的ASCII码，所想使其结果为6，应为int('6')-int('0'),或者采用哈希表来表示d，则无需转换。
2. 当字符为7或9时所对应的可选字符有4个

（1）C++——递归

class Solution {
public:
    vector letterCombinations(string digits) {
        vector res;
        int n = digits.length();
        if(n==0) return res;
        vector d {"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
        string temp="";
        help(digits,n,d,res,temp,0,0);
        help(digits,n,d,res,temp,0,1);
        help(digits,n,d,res,temp,0,2);
        if(digits[0]=='7' || digits[0]=='9')
            help(digits,n,d,res,temp,0,3);
        return res;
    }
    int help(string& digits,int n,vector& d,vector& res,string temp,int i,int j){
        temp += d[int(digits[i]-'0')][j];
        if(i==n-1){
            res.emplace_back(temp);
            return 0;
        }
        help(digits,n,d,res,temp,i+1,0);
        help(digits,n,d,res,temp,i+1,1);
        help(digits,n,d,res,temp,i+1,2);
        if(digits[i+1]=='7' || digits[i+1]=='9')
            help(digits,n,d,res,temp,i+1,3);
        return 0;
    }
};

（2）C+±-迭代法

class Solution {
public:
    vector letterCombinations(string digits) {
        vector res;
        if(digits.length() == 0) return res;
        res.push_back("");
        vector v = {"", "", "abc", "def", "ghi", "jkl", "mno","pqrs", "tuv", "wxyz"};
        for(int i = 0; i < digits.size(); i++) {
            string s = v[digits[i] - '0'];
            vector temp;
            for(int j = 0; j < s.length(); j++)
                for(int k = 0; k < res.size(); k++)
                    temp.push_back(res[k] + s[j]);
            res = temp;
        }
        return res;
    }
};

(3)python–递归

class Solution:
    def letterCombinations(self, digits: str) -> List[str]:
        n = len(digits)
        if n==0:
            return []
        d = {}
        d['2']='abc'
        d['3']='def'
        d['4']='ghi'
        d['5']='jkl'
        d['6']='mno'
        d['7']='pqrs'
        d['8']='tuv'
        d['9']='wxyz'
        res = []
        temp = ''
        self.diedai(digits,n,d,temp,res,0,0)
        self.diedai(digits,n,d,temp,res,0,1)
        self.diedai(digits,n,d,temp,res,0,2)
        if(digits[0] in '79'):
            self.diedai(digits,n,d,temp,res,0,3) 
        return res

    def diedai(self,digits,n,d,temp,res,i,j):
        temp += d[digits[i]][j]
        if i==n-1:
            res.append(temp)
            return 
        self.diedai(digits,n,d,temp,res,i+1,0)
        self.diedai(digits,n,d,temp,res,i+1,1)
        self.diedai(digits,n,d,temp,res,i+1,2)
        if(digits[i+1] in '79'):
            self.diedai(digits,n,d,temp,res,i+1,3)

(4)python-迭代

class Solution:
    def letterCombinations(self, digits: str) -> List[str]:
        n = len(digits)
        if n==0:
            return []
        d = ["", "", "abc", "def", "ghi", "jkl", "mno","pqrs", "tuv", "wxyz"]
        res = ['']
        for each in digits:
            temp=[]
            for a in d[int(each)-int('0')]:
                for it in res:
                    it+=a
                    temp.append(it)
            res = temp
        return res