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程式語言 - LeetCode - C - 17. Letter Combinations of a Phone Number

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參考資訊：
https://www.cnblogs.com/grandyang/p/4452220.html

題目：


解答：

int dfs(const char *digits, int pos, int *cidx, char *cbuf, int *ridx, char **rbuf)
{
    int i = 0;
    const char *BTN[] = {
        "abc", "def",
        "ghi", "jkl", "mno",
        "pqrs", "tuv", "wxyz"
    };
 
    if (pos >= strlen(digits)) {
        strcpy(rbuf[(*ridx)++], cbuf);
 
        *cidx -= 1;
        cbuf[*cidx] = 0;
        return 0;
    }
 
    const char *t = BTN[digits[pos] - '2'];
    int len = strlen(t);
 
    for (i = 0; i < len; i++) {
        cbuf[(*cidx)++] = t[i];
        dfs(digits, pos + 1, cidx, cbuf, ridx, rbuf);
    }
    *cidx -= 1;
 
    return 0;
}
 
/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
char** letterCombinations(char *digits, int *returnSize)
{
    int i = 0;
    char **rbuf = NULL;
    char cbuf[256] = { 0 };
 
    rbuf = calloc(256, sizeof(char *));
    for (i = 0; i < 256; i++) {
        rbuf[i] = calloc(256, sizeof(char));
    }
 
    *returnSize = 0;
    if (strlen(digits) == 0) {
        return rbuf;
    }
 
    i = 0;
    dfs(digits, 0, &i, cbuf, returnSize, rbuf);
 
    return rbuf;
}