【LeetCode】17. *的字母组合

链接:

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

描述:

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。
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。

示例:
输入:"23"
输出:["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

说明:
尽管上面的答案是按字典序排列的,但是你可以任意选择答案输出的顺序。

vector<string> letterCombinations(string digits) {}

思路:回溯法

使用回溯法,穷举所有可能的情况来得到解。

当遍历完所有数字时,产生结果
当探索到某一个数字时,遍历数字对应的所有字符:
 当前状态添加一个字符
 探索下一个数字
 还原当前状态

C++

展开后查看
class Solution {
public:
    map<char, string> letterMap = {
        {'2', "abc"}, {'3', "def"}, {'4', "ghi"}, {'5', "jkl"}, 
        {'6', "mno"}, {'7', "pqrs"}, {'8', "tuv"}, {'9', "wxyz"}
    };
    vector<string> result;
    string current;
    void backTrack(string digits, int index){
        if(index == digits.size()){
            result.push_back(current);
            return;
        }
        char digit = digits[index];
        string letters = letterMap[digit];
        for(int i = 0; i < letters.size(); i++){
            current.push_back(letters[i]);
            backTrack(digits, index + 1);
            current.pop_back();
        }
    }
    vector<string> letterCombinations(string digits) {
        if(digits.size() != 0){
            backTrack(digits, 0);
        }
        return result;
    }
};

Java

展开后查看
class Solution {
    Map<Character, String> letterMap = new HashMap<>(){
        {
            put('2', "abc");
            put('3', "def");
            put('4', "ghi");
            put('5', "jkl");
            put('6', "mno");
            put('7', "pqrs");
            put('8', "tuv");
            put('9', "wxyz");
        }
    };
    List<String> result = new ArrayList<>();
    StringBuilder current = new StringBuilder().append("");
    void backTrack(String digits, int index){
        if(index == digits.length()){
            result.add(current.toString());
            return;
        }
        Character digit = digits.charAt(index);
        String letters = letterMap.get(digit);
        for(int i = 0; i < letters.length(); i++){
            current.append(letters.charAt(i));
            backTrack(digits, index + 1);
            current.deleteCharAt(current.length() - 1);
        }
    }
    public List<String> letterCombinations(String digits) {
        if(digits.length() == 0){
            return result;
        }
        backTrack(digits, 0);
        return result;
    }
}
原文地址:https://www.cnblogs.com/crazyBlogs/p/13159065.html