上一題：LeetCode第16題: threeSumClosest(C語(yǔ)言）
思路：如果第一個(gè)輸入的數(shù)字是1羽历，其對(duì)應(yīng)的字母為‘a(chǎn)bc’炼团，由于1對(duì)應(yīng)的字母有三個(gè)褥琐，則在輸出結(jié)果中分別添加三個(gè)字符串，每個(gè)字符串分別為‘a(chǎn)’,‘b’,‘c’;
如果第二個(gè)輸入的數(shù)字是7驱富，其對(duì)應(yīng)的字母為‘pqrs’,由于7對(duì)應(yīng)的字母有四個(gè)匹舞，所以輸出的結(jié)果數(shù)量有排列組合可知：3乘以4=12個(gè)結(jié)果，所以需要把上一步的結(jié)果復(fù)制成12份叫榕，然后將本次的‘pqrs’均勻填入12份結(jié)果中姊舵。

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;
}

本系列文章尖昏，旨在打造LeetCode題目解題方法构资，幫助和引導(dǎo)同學(xué)們開(kāi)闊學(xué)習(xí)算法思路，由于個(gè)人能力和精力的局限性迹淌，也會(huì)參考其他網(wǎng)站的代碼和思路己单，如有侵權(quán)，請(qǐng)聯(lián)系本人刪除句携。