leetcode 17 电话号码 C语言实现

测评

题目

leetcode 17
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。

void dfs(char *digits, char *str, int ind, char letter[][5], char **res, int *returnSize) {
    if (digits[ind] == '\0') {
        str[ind] = '\0';
        strcpy(res[*returnSize], str);
        (*returnSize)++;//谨记*的符号低于++，记得加括号
        return ;
    }
    int d = digits[ind] - '0', i = 0;
    while (letter[d][i] != '\0') {
        str[ind] = letter[d][i];
        dfs(digits, str, ind + 1, letter, res, returnSize);
        i++;
    }
}

char ** letterCombinations(char * digits, int* returnSize){
    *returnSize = 0;
    if (digits[0] == '\0') return digits;
    char letter[][5] = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
    int len = strlen(digits);
    int count = 1;
    for (int i = 0; i < len; i++) {
        count *= 4;
    }
    char **res = (char **)malloc(sizeof(char *) * count);
    for (int i = 0; i < count; i++) {
        res[i] = (char *)malloc(sizeof(char) * (len + 1));
        res[i][len] = '\0';
    }
    char *str = (char *)malloc(sizeof(char) * (len + 1));
    dfs(digits, str, 0, letter, res, returnSize);
    return res;
}