WA：30.串联所有单词的子串

仅能过 words 中不存在重复字符串的测试用例 😖

void nex(int *next, char *s)
{
    next[0] = 0;
    int now = 0, j = 1, lens = strlen(s);
    while (j < lens)
    {
        if (s[now] == s[j])
            next[j++] = ++now;
        else if (now)
            now = next[now - 1];
        else
            next[j++] = 0;
    }
}

int *findSubstring(char *s, char **words, int wordsSize, int *returnSize)
{
    int wordlen = strlen(words[0]), lens = strlen(s), has = wordsSize * (wordsSize + 1) * (2 * wordsSize + 1) / 6, i = 0;
    int *next = (int *)malloc(sizeof(int) * wordlen), *t = (int *)malloc(sizeof(int) * lens), *result = (int *)malloc(sizeof(int) * lens);
    int j = lens;
    while (j--)
    {
        t[j] = 0;
        result[j] = -1;
    }
    while (i < wordsSize)
    {
        nex(next, words[i]);
        int now = 0, s1 = 0;
        while (s1 < lens)
        {
            if (words[i][now] == s[s1])
            {
                now += 1;
                s1 += 1;
            }
            else if (now)
                now = next[now - 1];
            else
                s1 += 1;
            if (now == wordlen)
            {
                t[s1 - now] = (i + 1) * (i + 1);
                now = next[now - 1];
            }
        }
        i += 1;
    }
    i = 0;
    int answernum = 0;
    while (i < lens)
    {
        int hanow = 0;
        if (t[i])
        {
            hanow += t[i];
            int z = i + wordlen, step = 1;
            while (step < wordsSize && z < lens && t[z])
            {
                hanow += t[z];
                z += wordlen;
                step += 1;
            }
            if (hanow == has)
            {
                result[answernum] = i;
                answernum += 1;
            }
        }
        i += 1;
    }
    *returnSize = answernum;
    if (answernum == 0)
        return NULL;
    int n = 0;
    int *res = (int *)malloc(sizeof(int) * answernum);
    while (answernum--)
    {
        res[n] = result[n];
        n += 1;
    }
    return res;
}