LeetCode17. Letter Combinations of a Phone Number（C++）

Given a string containing digits from 2-9 inclusive, return all possible letter combinations that the number could represent.

A mapping of digit to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.

Example:

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

Note:

Although the above answer is in lexicographical order, your answer could be in any order you want.

解题思路：暴力枚举法

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

优化上述方法，由于需要将temp数组复制到result，耗费了大量时间，所以直接在result数组进行变化可以节省大量时间，需要记下的是erase的使用，直接括号内为迭代器。

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

方法二：DFS遍历

这里值得注意的是动态数组的初始化

string *a=new string[10]{"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};

另外string类型也可以使用push_back()和pop_back()

class Solution {
public:
    vector<string>result;
    string s,temp="";
    string *a=new string[10]{"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
    int len;
    vector<string> letterCombinations(string digits) {
        s=digits;
        len=digits.length();
        if(len==0)
            return result;
        dfs(0);
        return result;
    }
    void dfs(int index){
        if(index==len){
            result.push_back(temp);
            return;
        }
        for(int i=0;i<a[s[index]-'0'].length();i++){
            temp+=a[s[index]-'0'][i];
            dfs(index+1);
            temp.pop_back();
        }
    }
};