【重点】【回溯】17.电话号码的组合

原创已于 2025-08-16 02:33:45 修改 · 419 阅读

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#深度优先 #dfs

于 2023-12-18 19:33:52 首次发布

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文章描述了一个使用深度优先搜索(DFS)算法解决的问题，将给定的数字字符串转换为其可能的字母组合。方法中定义了`letterCombinations`函数，通过递归调用`dfs`函数实现组合生成。

题目

Python

class Solution:
    def letterCombinations(self, digits: str) -> List[str]:
        if len(digits) == 0:
            return []
        num_dict = {
            '2': ['a', 'b', 'c'],
            '3': ['d', 'e', 'f'],
            '4': ['g', 'h', 'i'],
            '5': ['j', 'k', 'l'],
            '6': ['m', 'n', 'o'],
            '7': ['p', 'q', 'r', 's'],
            '8': ['t', 'u', 'v'],
            '9': ['w', 'x', 'y', 'z'],
        }
        
        res, tmp = [], []
        self.dfs(digits, tmp, num_dict, res)
        return res
    
    def dfs(self, digits, tmp, num_dict, res):
        if len(tmp) == len(digits):
            res.append(''.join(tmp))
            return
        
        start_idx = len(tmp)
        cur_digit = digits[start_idx]
        for i in range(len(num_dict[cur_digit])):
            tmp.append(num_dict[cur_digit][i])
            self.dfs(digits, tmp, num_dict, res)
            tmp.pop()  # list删除最后1个元素用pop方法

Java

法1：DFS

class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> res = new ArrayList<>();
        if (digits.length() == 0) {
            return res;
        }
        List<Integer> digitList = new ArrayList<>();
        for (int i = 0; i < digits.length(); ++i) {
            digitList.add(digits.charAt(i) - '0');
        }
        Map<Integer, char[]> map = new HashMap<>();
        map.put(2, new char[]{'a', 'b', 'c'});
        map.put(3, new char[]{'d', 'e', 'f'});
        map.put(4, new char[]{'g', 'h', 'i'});
        map.put(5, new char[]{'j', 'k', 'l'});
        map.put(6, new char[]{'m', 'n', 'o'});
        map.put(7, new char[]{'p', 'q', 'r', 's'});
        map.put(8, new char[]{'t', 'u', 'v'});
        map.put(9, new char[]{'w', 'x', 'y', 'z'});
        char[] tmp = new char[digitList.size()];
        dfs(map, digitList, 0, tmp, res);

        return res;
    }

    public void dfs(Map<Integer, char[]> map, List<Integer> digitList, 
                    int index, char[] tmp, List<String> res) {
        if (index == digitList.size()) {
            res.add(String.valueOf(tmp));
            return;
        }
        char[] charArray = map.get(digitList.get(index));
        int curSize = charArray.length;
        for (int i = 0; i < curSize; ++i) {
            tmp[index] = charArray[i];
            dfs(map, digitList, index + 1, tmp, res);
        }
    }
}