上一题：LeetCode第16题: threeSumClosest(C语言）
思路：如果第一个输入的数字是1，其对应的字母为‘abc’，由于1对应的字母有三个，则在输出结果中分别添加三个字符串，每个字符串分别为‘a’,‘b’,‘c’;
如果第二个输入的数字是7，其对应的字母为‘pqrs’,由于7对应的字母有四个，所以输出的结果数量有排列组合可知：3乘以4=12个结果，所以需要把上一步的结果复制成12份，然后将本次的‘pqrs’均匀填入12份结果中。

char** letterCombinations(char* digits, int* returnSize) {
    char *base[10] = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
    int length = strlen(digits);
    
    if(length == 0){
        *returnSize = 0;
        return '\0';
    }
    int max_result_count = 1;
    for(int i = 0; i < length; i++){
        max_result_count *= 4;
    }
    char **result = (char *)malloc(max_result_count * sizeof(char *));
    
    int real_count = 0;
    int letters_count = 0;
    for(int i = 0; i < length; i++){
        int num = digits[i] - 48;
        char *letters = base[num];
        if(num == 7 || num == 9){
            letters_count = 4;
        }
        else{
            letters_count = 3;
        }
        if(real_count == 0){
            for(int j = 0; j < letters_count; j++){
                char *new_result = (char *)malloc((length + 1) * sizeof(char));
                new_result[0] = letters[j];
                result[real_count++] = new_result; 
            }
        }
        else{
            int last_result_count = real_count;
            for(int j = last_result_count; j < last_result_count * letters_count; j++){
                int origin_index = j % last_result_count;
                char *origin_result = result[origin_index];
                char *new_result = (char *)malloc((length + 1) * sizeof(char));
                for(int k = 0; k < i; k++){
                    new_result[k] = origin_result[k];
                }
                result[real_count++] = new_result; 
            }
            
            for(int j = 0; j < real_count; j++){
                char *origin_result = result[j];
                int index = j / last_result_count;
                origin_result[i] = letters[index];
            }
        }
    }
    
    for(int j = 0; j < real_count; j++){
        char *sub_result = result[j];
        sub_result[length] = '\0';
    }
    
    *returnSize = real_count;
    return result;
}

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