【leetcode】特殊等价字符串组

 

int cmpChar(const void *a, const void *b) {
    return *(char *)a - *(char *)b;
}

int cmpStr(const void *a, const void *b) {
    return strcmp(*(char **)a, *(char **)b);
}

int numSpecialEquivGroups(char ** A, int ASize){
    for (int i = 0; i < ASize; i++) {
        int len = strlen(A[i]);
        // 偶数多后退1，奇数不用
        for (int idx1 = 1, idx2 = len - 1 - (len % 2 == 0); idx1 < idx2; idx1 += 2, idx2 -= 2) {
            char tmp = A[i][idx1];
            A[i][idx1] = A[i][idx2];
            A[i][idx2] = tmp;
        }
        // 分别排序前后两个部分；奇数与偶数长度时奇数个数是len/2，因此偶数个数len-odds
        int odds = len / 2, evens = len - odds;
        qsort(A[i], evens, sizeof(A[i][0]), cmpChar);
        qsort(A[i] + evens, odds, sizeof(A[i][evens]), cmpChar);
    }
    qsort(A, ASize, sizeof(A[0]), cmpStr);
    // 2. 暴力对比
    int cnt = 1;  // 肯定存在一个，向前向后比较都可以
    for (int i = 1; i < ASize; i++) {
        if (strcmp(A[i], A[i-1]) != 0) {
            cnt++;
        }
    }
    return cnt;
}
// ["abcd","cdab","cbad","xyzz","zzxy","zzyx"]

 

/*异或加上strchr*/
int numSpecialEquivGroups(char ** A, int ASize){
    int i,j,k,pair=0,opst=0,epst=0;
    int len = strlen(A[0]);
    int val1,val2;
    
    for (i=0; i<ASize; i++)
    {
        if (!A[i]) continue;
         char evenStr[20]={0};
         char oodStr[20]={0};
        opst=epst=0;
        for (k=0; k<len; k++)
        {
            if (k%2) {
                oodStr[k/2]=A[i][k];
                opst ^= A[i][k];
            }
            else{
                evenStr[k/2]=A[i][k];
                epst ^= A[i][k];
            }
        }
        for (j=i+1; j<ASize; j++)
        {
            val1=opst;
            val2=epst;
            if (!A[j]) continue;
            for (k=0; k<len; k++)
            {
                 if (k%2 && !strchr(oodStr,A[j][k]) || k%2==0 && !strchr(evenStr,A[j][k])) break;
                if (k%2)
                    val1 ^= A[j][k];
                else  
                    val2^= A[j][k];
            }
            if (k==len && !val1 && !val2)
                A[j]=NULL;
        }
        pair++;
    }
    return pair;
}