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第十七题：电话号码的字母组合

article 2024/9/23 17:21:38

题目描述

给定一个仅包含数字 2-9 的字符串，返回所有可能的由它组成的字母组合。你可以假设输入字符串至少包含一个数字，并且不超过3位数字。

实现思路

使用哈希表或数组存储每个数字对应的字符，然后通过递归或迭代的方式生成所有可能的组合。如果字符串长度为n，则可以看作是n层循环，每层循环可以选择对应数字的所有字符之一。

算法实现

C语言实现

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void backtrack(char *digits, int pos, char *combination, int *resultLength, char **result) {
    static const char *table[] = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
    if (pos == strlen(digits)) {
        result[*resultLength] = malloc(strlen(combination) + 1);
        strcpy(result[*resultLength], combination);
        (*resultLength)++;
    } else {
        for (int i = 0; table[digits[pos] - '0'][i]; ++i) {
            combination[pos] = table[digits[pos] - '0'][i];
            backtrack(digits, pos + 1, combination, resultLength, result);
        }
    }
}

int* letterCombinations(char *digits, int *returnSize) {
    if (!strlen(digits)) return NULL;
    int resultLength = 0;
    char *combination[100] = {0};
    *returnSize = 0;
    backtrack(digits, 0, (char[5]){0}, &resultLength, combination);
    int *result = malloc(resultLength * sizeof(int));
    for (int i = 0; i < resultLength; ++i) {
        result[i] = (int)combination[i];
    }
    return result;
}

Python实现

class Solution:
    def letterCombinations(self, digits: str) -> List[str]:
        if not digits:
            return []
        
        phoneMap = {'2': 'abc', '3': 'def', '4': 'ghi', '5': 'jkl', 
                    '6': 'mno', '7': 'pqrs', '8': 'tuv', '9': 'wxyz'}
        def backtrack(combination, next_digits):
            if not next_digits:
                combinations.append(combination)
            else:
                for letter in phoneMap[next_digits[0]]:
                    backtrack(combination + letter, next_digits[1:])
        
        combinations = []
        backtrack("", digits)
        return combinations

Java实现

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

public class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> combinations = new ArrayList<>();
        if (digits.length() == 0) {
            return combinations;
        }
        Map<Character, String> phoneMap = new HashMap<>();
        phoneMap.put('2', "abc"); phoneMap.put('3', "def"); phoneMap.put('4', "ghi");
        phoneMap.put('5', "jkl"); phoneMap.put('6', "mno"); phoneMap.put('7', "pqrs");
        phoneMap.put('8', "tuv"); phoneMap.put('9', "wxyz");

        backtrack(combinations, phoneMap, digits, 0, new StringBuilder());
        return combinations;
    }

    private void backtrack(List<String> combinations, Map<Character, String> phoneMap, String digits, int index, StringBuilder combination) {
        if (index == digits.length()) {
            combinations.add(combination.toString());
        } else {
            String letters = phoneMap.get(digits.charAt(index));
            for (int i = 0; i < letters.length(); i++) {
                combination.append(letters.charAt(i));
                backtrack(combinations, phoneMap, digits, index + 1, combination);
                combination.deleteCharAt(index);
            }
        }
    }
}