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 {
2 public:
3 map<char,string> m;
4 int length;
5 vector<string> result;
6 vector<string> letterCombinations(string digits) {
7 m['2']="abc";
8 m['3']="def";
9 m['4']="ghi";
10 m['5']="jkl";
11 m['6']="mno";
12 m['7']="pqrs";
13 m['8']="tuv";
14 m['9']="wxyz";
15 int n=digits.length();
16 // vector<string> a;
17 // if(n==0) return a;
18 length=n;
19 string re="";
20 letterComb(digits,m,0,re);
21 return result;
22 }
23 void letterComb(string &digits,map<char,string> & m,int n,string re) //这里用的值传递
24 {
25 if(n==length)
26 {
27 result.push_back(re);
28 return;
29 }
30 string tmp=m[digits[n]];
31 int l=tmp.length();
32 for(int i=0;i<l;i++)
33 {
34 char c=tmp[i];
35 letterComb(digits,m,n+1,re+c); //string 类型可以加上char类型哦
36 }
37 }
38
39 };
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：
1 class Solution {
2 public:
3 vector<string> result;
4 int length;
5 string m[8]={"abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
6
7 vector<string> letterCombinations(string digits) {
8 int n=digits.length();
9 length=n;
10 string re="";
11 letterComb(digits,0,re);
12 return result;
13 }
14 void letterComb(string &digits,int n,string &re)
15 {
16 if(n==length)
17 {
18 result.push_back(re);
19 return;
20 }
21 string tmp=m[digits[n]-'2'];
22 int l=tmp.length();
23 for(int i=0;i<l;i++)
24 {
25 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
26 letterComb(digits,n+1,re);
27 re.pop_back();
28 }
29 }
30
31 };
方法二：还有个不递归的方法
1 class Solution {
2 public:
3 vector<string> letterCombinations(string digits) {
4 int n=digits.size();
5 vector<string> result(1,"");
6 string m[8]={"abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
7 for(int i=0;i<n;i++)
8 {
9 string s=m[digits[i]-'2'];
10 int l=s.size();
11 int size=result.size();
11 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
12 {
13 for(int k=0;k<l;k++)
14 {
15 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
16 {
17 result[j]+=s[k];
18 }
19 else
20 {
21 result.push_back(result[j]+s[k]);
22 }
23 }
24 }
25 }
26 return result;
27
28 }
29
30 };
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