【字符指针等_例题详解】

void young_tableau_search(int arr[3][3], int k, int* px, int* py)
{
    int x = 0;
    int y = *py-1;
    int flag = 0;

    while (x <= *px - 1 && y >= 0)
    {
        if (arr[x][y] < k)
        {
            x++;
        }
        else if (arr[x][y] > k)
        {
            y--;
        }
        else
        {
            *px = x;
            *py = y;
            return;
        }
    }
    *px = -1;
    *py = -1;

}

int main()
{
    int arr[3][3] = { 1,2,3,4,5,6,7,8,9 };
    int k = 0;
    scanf("%d", &k);
    int x = 3;
    int y = 3;
    young_tableau_search(arr, k, &x, &y);
    if (x == -1 && y == -1)
    
        printf("没找到\n");
   else
       printf("找到了，下标是:%d %d\n", x, y);

    return 0;
}

void reverse(char* left, char* right)
{
    assert(right && left);
    while (left < right)
    {
        char tmp = *left;
        *left = *right;
        *right = tmp;
        left++;
        right--;
    }
}

void left_move(char* str, int k)
{
    int len = strlen(str);
    k %= len;
    //左边逆序
    reverse(str, str + k - 1);
    //右边逆序
    reverse(str + k, str + len - 1);
    //整体逆序
    reverse(str, str + len - 1);
}

int main ()
{
    char arr[] = "abcdef";
    int k = 0;
    scanf("%d", &k);
    left_move(arr, k);
    printf("%s\n", arr);
    return 0;
}

//判断一字符串是否为另外一个字符串旋转之后的字符串
//方法1
int is_left_move(char* str1, char* str2)
{
    int j = 0;
    int len = strlen(str1);
    for (j = 0; j < len; j++)
    {
        char tmp = *str1;
        int len = strlen(str1);
        int i = 0;
        for (i = 0; i < len - 1; i++)
        {
            *(str1 + i) = *(str1 + i + 1);
        }
        *(str1 + len - 1) = tmp;
        if (strcmp(str1, str2) == 0)
            return 1;
    }
}

int main()
{
    char arr1[] = "abcdef";
    char arr2[] = "cdefab";

    int ret = is_left_move(arr1, arr2);
    if (ret == 1)
        printf("Yse\n");
    else
        printf("NO\n");

    return 0;
}