Leetcode: Letter Combinations of a Phone Number

简单DFS

总结

第一次提交忘了 clear 向量

digits[i]-'0'-2 这种写法 work

#include <iostream>
#include <vector>
#include <string>
using namespace std;
vector<vector<char> >phone;
vector<string> result;
class Solution {
public:
    void init() {
        int count = 0;
        for(int i = 1; i <= 8; i ++) {
            vector<char> tmp;
            for(int j = 0; j < 3; j ++) {
                tmp.push_back('a'+count);
                count++;
            }
            if(i == 6 || i == 8) {
                tmp.push_back('a'+count);
                count++;
            }
            phone.push_back(tmp);
        }
    }
    void proceeds(string digits, const int &i, const int &n, string partial) {
        if(i == n) {
            result.push_back(partial);
            return;
        }
        //cout << digits[i]-'0'-2 << endl;
        for(int j = 0; j < phone[digits[i]-'0'-2].size(); j ++) {
            proceeds(digits, i+1, n, partial+ phone[digits[i]-'0'-2][j]);
        }

    }
    vector<string> letterCombinations(string digits) {
        result.clear();
        phone.clear();
        init();
        proceeds(digits, 0, digits.size(), "");
        return result;
    }
};