Letter Combinations of a Phone Number

思路很简单，就是通过map对应，随后通过递归调用，这里遇到了个问题就是开始用string &re ,然后往里传re+c是不可以的，本来想省点内存，结果不行，所以只能值传递了，后来才知道string
竟然还有push_back()和pop_back(),服了。
 1 class Solution {
public:
    map<char,string> m;
    int length;
    vector<string> result;
    vector<string> letterCombinations(string digits) {
        m['2']="abc";
        m['3']="def";
        m['4']="ghi";
        m['5']="jkl";
        m['6']="mno";
        m['7']="pqrs";
        m['8']="tuv";
        m['9']="wxyz";
    int n=digits.length();
    // vector<string> a;
    // if(n==0) return a;
    length=n;
    string re="";
    letterComb(digits,m,0,re);
    return result;
    }
   void letterComb(string &digits,map<char,string> & m,int n,string re)               //这里用的值传递
   {
         if(n==length)
        {
             result.push_back(re);
            return;
        }
        string tmp=m[digits[n]];
        int l=tmp.length();
        for(int i=0;i<l;i++)
        {
              char c=tmp[i];
              letterComb(digits,m,n+1,re+c);                                     //string 类型可以加上char类型哦
        }
   }

};

这里map还可以用

string str[8]={"abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};替代，只是注意每次要减‘2’，
string tmp=str[digits[i]-'2']
或者用 string strT[10] = {"","","abc","def","ghi","jkl","mno","qprs","tuv","wxyz"}; 这样是对应的，递归方法相同
修改后方法如下：

class Solution {
public:
    vector<string> result;
    int length;
    string m[8]={"abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
    
    vector<string> letterCombinations(string digits) {
    int n=digits.length();
    length=n;
    string re="";
    letterComb(digits,0,re);
    return result;
    }
   void letterComb(string &digits,int n,string &re)
   {
         if(n==length)
        {
             result.push_back(re);
             return;
        }
        string tmp=m[digits[n]-'2'];
        int l=tmp.length();
        for(int i=0;i<l;i++)
        {
              re.push_back(tmp[i]);               //这里string用push_back()和pop_back()，我也是第一次发现别人能这么用，厉害啊
              letterComb(digits,n+1,re);
              re.pop_back();
        }
   }

};

 

方法二：还有个不递归的方法

class Solution {
public:
    vector<string> letterCombinations(string digits) {
    int n=digits.size();
    vector<string> result(1,"");
    string m[8]={"abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
    for(int i=0;i<n;i++)
    {
        string s=m[digits[i]-'2'];
        int l=s.size();
11           int size=result.size();
        for(int j=0;j<=size-1;j++)             //一定注意这里，必须先取得result.size()，否则会造成死循环
        {
            for(int k=0;k<l;k++)
            {
               if(k==l-1)            //当是最后一个时，就不能再push啦，不然就重了，要把自己改了
               {
                  result[j]+=s[k]; 
               }
               else
               {
                   result.push_back(result[j]+s[k]);
               }
            }
        }
    }
    return result;
    
    }

};