【leetcode】两个列表的最小索引总和

/*双循环暴力解法*/
char ** findRestaurant(char ** list1, int list1Size, char ** list2, int list2Size, int* returnSize){
    int i,j,n=0,pst=0,min=2000;
    char** arr = (char**)calloc(1000,sizeof(char*));
    for (i=0; i<list1Size; i++)
        for (j=0; j<list2Size; j++)
            if (!strcmp(list1[i],list2[j]) && i+j<=min)
            {
                if (i+j<min){
                    pst=n;
                    min = i+j;
                }
                arr[n++] = list1[i];
                break;
            }    
    *returnSize = n-pst;
    return &arr[pst];
}

/*对字符串每个字母按顺序快排，结构体存放字符串地址和原来在数组的索引*/
typedef struct _Data{
    int index;
    char *info;
}Data;

int cmp(const void *a, const void *b){  
    int ret = strcmp(((Data*)a)->info, ((Data*)b)->info);    
    return ret;
}

char ** findRestaurant(char ** list1, int list1Size, char ** list2, int list2Size, int* returnSize){
    *returnSize = 0;
    char** res = (char**)malloc((list1Size > list2Size ? list2Size : list1Size) * sizeof(char*));
    Data *src1 = (Data*)malloc(list1Size * sizeof(Data));
    Data *src2 = (Data*)malloc(list2Size * sizeof(Data));
    int i, j = 0, val, max = list1Size + list2Size, sum;
    for (i = 0; i < list1Size; ++i) {
        src1[i].index = i;
        src1[i].info = list1[i];
    }
    for (i = 0; i < list2Size; ++i){
        src2[i].index = i;
        src2[i].info = list2[i];
    }
    qsort(src1, list1Size, sizeof(Data), cmp);
    qsort(src2, list2Size, sizeof(Data), cmp);
    
    i = 0, j = 0;
    while (i < list1Size && j < list2Size){
        val = strcmp(src1[i].info, src2[j].info);
        if (val < 0){
            ++i;
        } else if (val == 0){
            sum = src1[i].index + src2[j].index;
            if (sum < max){
                max = sum;
                *returnSize = 0;
                res[(*returnSize)++] = src1[i].info;
            } else if (sum == max){
                res[(*returnSize)++] = src1[i].info;
            }
            ++i;
            ++j;
        } else {
            ++j;
        }
    }
    free(src1);
    free(src2);
    return res;
}