Character Pointer Scrolling Record

1. The following description of "pointer" is incorrect:

2. The following code description is correct:

char* p = "hello beauty!";
const char* p = "hello beauty!";//More appropriate writing

3. The correct description of the array pointer is:

4. Which of the following is the array pointer:

The array pointer is a pointer to the array

Pointer array is an array, which stores pointers

5. Which of the following codes is wrong?

int main()
{
  int *p = NULL;
  int arr[10] = {0};//arr is the array name, and the array name is the address int of the first element*
 //int* p = arr;
  return 0;
}

A.p = arr;
B.int (*ptr)[10] = &arr;//Pointer array
C.p = &arr[0];//Equivalent to A
D.p = &arr;//Take the array address and put it into the integer pointer. The two do not match

6. The following code describes the array name incorrectly:

int main()
{
  int arr[10] = {0};
  return 0;
}

A.Array name arr and&arr It's the same//&The values of arr and arr are the same, but their meanings are different
B.sizeof(arr)，arr Represents the entire array
C.&arr，arr Represents the entire array
D.except sizeof(arr)and&arr Array name in, array name in other places arr，Are the addresses of the first element of the array.

7. Define array:

//How to define an int pointer array with 10 array elements:
int * arr[10];
A.int a[10]//integer array 
B.int (*a)[10]//Array pointer
C.int *a[10];
D.int (*a[10])(int);//Function pointer array

8. The execution result of the following code is

int main()
{
  char str1[] = "hello bit.";
  char str2[] = "hello bit.";
  char *str3 = "hello bit.";
  char *str4 = "hello bit.";
  if(str1 == str2)
    printf("str1 and str2 are same\n");
  else
    printf("str1 and str2 are not same\n");//Print
  if(str3 == str4)
    printf("str3 and str4 are same\n");//Print
  else
    printf("str3 and str4 are not same\n");
  return 0; 
}

9. Young matrix

There is a number matrix. Each row of the matrix is increasing from left to right, and the matrix is increasing from top to bottom. Please write a program to find whether a number exists in such a matrix.

Requirement: time complexity is less than O(N) ---- avoid traversing data

1 2 3

4 5 6

7 8 9

Thought expansion

code implementation

void find_num_in_young (int arr[3][3],int k,int *px,int *py)
{
	int i = 0;
	int j =*py - 1;
	int flag = 0;
	while (i <=* px - 1 && j >=0)
	{
		if (arr[i][j] < k)
			i++;
		else if (arr[i][j] > k)
			j--;
		else
		{
			//eureka
			flag = 1;
			*px = i;
			*py = j;
			break;
		}
	}
	if (flag == 0)
	{
		*px = -1;
		*py = -1;
	}
}

Principal function

int main()
{
	int arr[3][3] = { 1,2,3,4,5,6,7,8,9 };
	int k=0;
	scanf("%d", &k);//Enter Search Number
	//Traversal array
	int x = 0;
	int y = 3;
	//Return type parameter
	find_num_in_young(arr, k, &x, &y);
	if (x == 1 && y == -1)
		printf("can't find\n");
	else
		printf("Yes, the subscript is:");
	return 0;
}

10. Left rotation of string

Implement a function that can rotate k characters in a left-handed string.

Code implementation 1

Disadvantages: Some codes will move repeatedly

ABCD Rotate a character to the left to get BCDA

ABCD Rotate left two characters to get CDAB

void left_move(char arr[], int k)
{
//Rotate only one character at a time, and execute this action k times
	int i = 0;
	int len = strlen(arr);
	k %= len;//Improve efficiency and keep cycle times in single digits
	for (i = 0; i < k; i++)
	{
		//1
		char tmp = arr[0];
		//2 Move all the characters behind one position forward
		int j = 0;
		for (j = 0; j < len - 1; j++)
			arr[j] = arr[j + 1];
		//3
		arr[len - 1] = tmp;
	}
}

Code implementation 2:

One code is exchanged only twice

void reverse(char* left, char* right)
{
	assert(left && right);
	while (left < right)
	{
		char tmp = *left;
		*left = *right;
		*right = tmp;
		left++;
		right--;
	}
}
void left_move(char arr[], int k)
{
	//Rotate only one character at a time, and execute this action k times
	int i = 0;
	int len = strlen(arr);
	k %=  len;
	reverse(arr, arr + k);
	reverse(arr+k,arr+len-1);
	reverse(arr, arr + len - 1);

}

Main function:

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

11. String rotation result

Write a function to determine whether one string is the string after another string is rotated.

For example, given s1=AABCD and s2=BCDAA, 1 is returned

Given s1=abcd and s2=ACBD, 0 is returned

AABCD Rotate a character to the left to get ABCDA

AABCD Rotate two characters left to get BCDAA

AABCD Rotate a character to the right to get DAABC

Code implementation 1

int is_left_move(char arr1[], char arr2[])
{
	int len = strlen(arr1);
	int i = 0;
	for (i = 0; i < len; i++)
	{
		char tmp = arr1[0];
		int j = 0;
		for (j = 0; j < len - 1; j++)
			arr1[j] = arr1[j + 1];
		arr1[len - 1] = tmp;
		if (strcmp(arr1, arr2) == 0)
			return 1;
	}
	return 0;
}

Code Implementation 2

int is_left_move(char arr1[], char arr2[])
{
	//strcat strstr strncat
	int len1 = strlen(arr1);
	int len2 = strlen(arr2);
	if (len1 != len2)
		return 0;
	strncat(arr1, arr1, len1);
	char* ret = strstr(arr1, arr2);
	if (ret == NULL)
		return 0;
	else
		return 1;
}

main function

//Write a function to determine whether a string is a string after rotation of another string
int main()
{
	char arr1[] = "ABCDEF";
    //char arr1[20] = "ABCDEF";// Note: Modern code 2 is implemented on the premise that the arr1 space needs to be large enough
    char arr2[] = "CDEFAB";
	int ret = is_left_move(arr1, arr2);//Judge whether arr2 is obtained by rotation of arr1
	if (1 == ret)
		printf("Yes\n");
	else
		printf("No\n");
	return 0;
}

``c
//Write a function to determine whether a string is a string after rotation of another string
int main()
{
char arr1[] = "ABCDEF";
//char arr1[20] = “ABCDEF”;// Note: Modern code 2 is implemented on the premise that the arr1 space needs to be large enough
char arr2[] = "CDEFAB";
int ret = is_left_move(arr1, arr2);// Judge whether arr2 is obtained by rotation of arr1
if (1 == ret)
printf("Yes\n");
else
printf("No\n");
return 0;
}