Leetcode:17. Letter Combinations of a Phone Number

Description

解题思路

1.DFS

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        vector<string> result;
        string save;
        if (digits.size() == 0) return vector<string>();
        int convert[10] = {0, 3, 6, 9, 12, 15, 19, 22, 26};
        dfs(result, save, digits, 0, digits.size(), convert);
        return result;
    }
    void dfs(vector<string> &str, string save, string dig, int now, int size, int *convert){
        if(now == size){
            str.push_back(save);
            return;
        }
        int tmp = dig[now] - '0' - 2;//change '2' to 0, '3' to 1...
        for(int i = convert[tmp]; i < convert[tmp+1]; i++){
            save.push_back('a' + i);//ex: a,b,c | d,e,f
            dfs(str, save, dig, now + 1, size, convert);
            save.pop_back();
        }
    }
};

2.采用迭代方法求解，以123为例

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        if (digits.size() == 0) return vector<string>();
        static const vector<string> v = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        vector<string> result;
        result.push_back("");
        for (int i = 0; i < digits.size(); i++) {
            int num = digits[i]-'0';
            if (num > 9 || num < 0) break;
            string s = v[num];
            vector<string> t;
            for (int j = 0; j < s.size(); j++) {
                for (int k = 0; k < result.size(); k++) {
                    t.push_back(result[k]+s[j]);
                }
            }
            result.swap(t);
        }
        return result;
    }
};