LeetCode 17. 电话号码的字母组合 中等

作者:Choi Yang

更新于:7 个月前

字数统计:1k 字

阅读时长:5 分钟

题目描述

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。

解题思路

采用回溯做法，对于当前选项，我们可以重复选择，所以 for 循环那里从 0 开始，对于字母组合我们做一个 map映射即可。

参考 xiao_ben_zhu 大佬的图解

javascript

var letterCombinations = function (digits) {
  if (!digits.length) return [];
  // 直接映射
  const map = {
    2: "abc",
    3: "def",
    4: "ghi",
    5: "jkl",
    6: "mno",
    7: "pqrs",
    8: "tuv",
    9: "wxyz",
  };
  let res = [];
  let dfs = (cur, start) => {
    if (start >= digits.length) {
      res.push(cur);
      return;
    }
    // 取当前可选的字母组合
    let str = map[digits[start]];
    for (let i = 0; i < str.length; i++) {
      dfs(cur + str[i], start + 1);
    }
  };
  dfs("", 0);
  return res;
};

cpp

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        if (digits.empty()) return {};
        vector<string> res;
        string cur;
        dfs(digits, 0, cur, res);
        return res;
    }
    void dfs(string& digits, int start, string& cur, vector<string>& res) {
        if (start >= digits.size()) {
            res.push_back(cur);
            return;
        }
        string str = map[digits[start]];
        for (int i = 0; i < str.size(); i++) {
            cur.push_back(str[i]);
            dfs(digits, start + 1, cur, res);
            cur.pop_back();
        }
    }
private:
    unordered_map<char, string> map = {
        {'2', "abc"},
        {'3', "def"},
        {'4', "ghi"},
        {'5', "jkl"},
        {'6', "mno"},
        {'7', "pqrs"},
        {'8', "tuv"},
        {'9', "wxyz"}
    };
};

java

class Solution {
    public List<String> letterCombinations(String digits) {
        if (digits.length() == 0) return new ArrayList<>();
        List<String> res = new ArrayList<>();
        dfs(digits, 0, new StringBuilder(), res);
        return res;
    }
    private void dfs(String digits, int start, StringBuilder cur, List<String> res) {
        if (start >= digits.length()) {
            res.add(cur.toString());
            return;
        }
        String str = map.get(digits.charAt(start));
        for (int i = 0; i < str.length(); i++) {
            cur.append(str.charAt(i));
            dfs(digits, start + 1, cur, res);
            cur.deleteCharAt(cur.length() - 1);
        }
    }
    private Map<Character, String> map = new HashMap<Character, String>() {
        {
            put('2', "abc");
            put('3', "def");
            put('4', "ghi");
            put('5', "jkl");
            put('6', "mno");
            put('7', "pqrs");
            put('8', "tuv");
            put('9', "wxyz");
        }
    };
}

python

class Solution:
    def letterCombinations(self, digits: str) -> List[str]:
        if not digits: return []
        res = []
        self.dfs(digits, 0, '', res)
        return res
    def dfs(self, digits, start, cur, res):
        if start >= len(digits):
            res.append(cur)
            return
        str = self.map[digits[start]]
        for i in range(len(str)):
            self.dfs(digits, start + 1, cur + str[i], res)
    map = {
        '2': 'abc',
        '3': 'def',
        '4': 'ghi',
        '5': 'jkl',
        '6': 'mno',
        '7': 'pqrs',
        '8': 'tuv',
        '9': 'wxyz'
    }

解法 2

这个是没用回溯之前写的一份代码，简单来说就是利用了层次遍历的特性，反正每次取字母都是可以重复的，直接遍历即可，然后进队列。

javascript

var letterCombinations = function (digits) {
  if (!digits.length) return [];
  const map = {
    2: "abc",
    3: "def",
    4: "ghi",
    5: "jkl",
    6: "mno",
    7: "pqrs",
    8: "tuv",
    9: "wxyz",
  };
  let queue = [];
  queue.push("");
  for (let i = 0; i < digits.length; i++) {
    let size = queue.length;
    while (size--) {
      let cur = queue.shift();
      let str = map[digits[i]];
      for (let j = 0; j < str.length; j++) {
        queue.push(cur + str[j]);
      }
    }
  }
  return queue;
};

cpp

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        if (digits.empty()) return {};
        vector<string> res;
        queue<string> q;
        q.push("");
        for (int i = 0; i < digits.size(); i++) {
            int size = q.size();
            while (size--) {
                string cur = q.front();
                q.pop();
                string str = map[digits[i]];
                for (int j = 0; j < str.size(); j++) {
                    q.push(cur + str[j]);
                }
            }
        }
        while (!q.empty()) {
            res.push_back(q.front());
            q.pop();
        }
        return res;
    }
private:
    unordered_map<char, string> map = {
        {'2', "abc"},
        {'3', "def"},
        {'4', "ghi"},
        {'5', "jkl"},
        {'6', "mno"},
        {'7', "pqrs"},
        {'8', "tuv"},
        {'9', "wxyz"}
    };
};

java

class Solution {
    public List<String> letterCombinations(String digits) {
        if (digits.length() == 0) return new ArrayList<>();
        List<String> res = new ArrayList<>();
        Queue<String> q = new LinkedList<>();
        q.offer("");
        for (int i = 0; i < digits.length(); i++) {
            int size = q.size();
            while (size-- > 0) {
                String cur = q.poll();
                String str = map.get(digits.charAt(i));
                for (int j = 0; j < str.length(); j++) {
                    q.offer(cur + str.charAt(j));
                }
            }
        }
        while (!q.isEmpty()) {
            res.add(q.poll());
        }
        return res;
    }
    private Map<Character, String> map = new HashMap<Character, String>() {
        {
            put('2', "abc");
            put('3', "def");
            put('4', "ghi");
            put('5', "jkl");
            put('6', "mno");
            put('7', "pqrs");
            put('8', "tuv");
            put('9', "wxyz");
        }
    };
}

python

class Solution:
    def letterCombinations(self, digits: str) -> List[str]:
        if not digits: return []
        res = []
        q = collections.deque()
        q.append('')
        for i in range(len(digits)):
            size = len(q)
            while size:
                cur = q.popleft()
                str = self.map[digits[i]]
                for j in range(len(str)):
                    q.append(cur + str[j])
                size -= 1
        while q:
            res.append(q.popleft())
        return res
    map = {
        '2': 'abc',
        '3': 'def',
        '4': 'ghi',
        '5': 'jkl',
        '6': 'mno',
        '7': 'pqrs',
        '8': 'tuv',
        '9': 'wxyz'
    }

javascript

学如逆水行舟，不进则退

Contributors

Choi Yang