【算法day16】电话号码的字母组合

电话号码的字母组合

给定一个仅包含数字 2-9 的字符串，返回所有它能表示的字母组合。答案可以按 任意顺序 返回。

给出数字到字母的映射如下（与电话按键相同）。注意 1 不对应任何字母。

https://leetcode.cn/problems/letter-combinations-of-a-phone-number/submissions/613453566/

DFS"

class Solution {
public:
    vector<string> phone_string = {"abc", "def",  "ghi", "jkl",
                                   "mno", "pqrs", "tuv", "wxyz"};
    string tmp;
    vector<string> ans;
    void DFS(int pos, string digits) {
        if (pos == digits.size()) {
            ans.push_back(tmp);
            return;
        }
        int num = digits[pos] - '0' - 2;
        for (int i = 0; i < phone_string[num].size(); i++) {
            tmp.push_back(phone_string[num][i]);
            DFS(pos + 1, digits);
            tmp.pop_back();
        }
    }
    vector<string> letterCombinations(string digits) {
        if (digits.size() == 0)
            return ans;
        DFS(0, digits);
        return ans;
    }
};

BFS

class Solution {
public:
    vector<string> phone_string = {"abc", "def",  "ghi", "jkl",
                                   "mno", "pqrs", "tuv", "wxyz"};
    queue<string> q;

    vector<string> BFS(vector<string>& ch) {
        vector<string> ans;
        for (int i = 0; i < ch[0].size(); i++) {
            string str(1, ch[0][i]);
            q.push(str);
        }
        int j = 1;
        while (q.front().size() < ch.size()) {
            int len = q.size();

            for (int k = 0; k < len; k++) {
                string top = q.front();
                q.pop();
                for (int i = 0; i < ch[j].size(); i++) {
                    q.push(top + ch[j][i]);
                }
            }
            j++;
        }
        for (; !q.empty(); q.pop()) {
            ans.push_back(q.front());
        }
        return ans;
    }
    vector<string> letterCombinations(string digits) {
        vector<string> ans;
        if (digits.size() == 0)
            return ans;
        vector<string> ch;
        for (int i = 0; i < digits.size(); i++) {
            ch.push_back(phone_string[digits[i] - '0' - 2]);
        }
        return BFS(ch);
    }
};