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C language double pointer -- classic question type
2022-07-06 08:32:00 【Finally - null】
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1. Deletes the specified number from the sequence
Link to Niuke :
describe
There is a sequence of integers ( There may be duplicate integers ), Now delete a specified integer , Output the sequence after deleting the specified number , There is no change in the front and back positions of the undeleted numbers in the sequence .
Data range : Both the length of the sequence and the values in the sequence satisfy 1≤n≤50
Input description :
Enter an integer in the first line (0≤N≤50).
Second line input N It's an integer , Enter space delimited N It's an integer .
On the third line, enter an integer you want to delete .
Output description :
Output as a line , Delete the sequence after the specified number .
Example 1
Input :
6
1 2 3 4 5 9
4
Output :1 2 3 5 9Example 2
Input :
5
1 2 3 4 6
5
Output :1 2 3 4 6
Ideas :
Define two variables , Are subscripted from the array 0 The position begins ,i Variables traverse the entire array ,j Variables are used to store elements that are not deleted , When i After finding the deleted element i Continue to visit backwards ,j No more , When it is not the element to be deleted , Mark the subscript as i The elements of are stored in j Subscript , then j Continue to add .
#include<stdio.h> int main() { int n = 0; int arr[50] = { 0 }; scanf("%d", &n); int i = 0; for (i = 0; i < n; i++) { scanf("%d", &arr[i]); } int del = 0; int j = 0; scanf("%d", &del); for (i = 0; i < n; i++) { if (arr[i] != del) { arr[j++] = arr[i]; } } for (i = 0; i < j; i++) { printf("%d ", arr[i]); } return 0; }
2. Remove duplicates from the sequence
Link to Niuke :
describe
Input n A sequence of integers , This sequence is required to be de duplicated . So called de duplication , It refers to every repeated integer in this sequence , Keep only the position where the number first appears , Delete the rest of the position .
Input description :
The input contains two lines , The first line contains a positive integer n(1 ≤ n ≤ 1000), Represents the number of numbers in the second row sequence ; The second line contains n It's an integer ( Range 1~5000), Separate... With spaces .
Output description :
Output as a line , Output the number after de duplication in the order of input , Separate... With spaces .
Example 1
Input :
5
10 12 93 12 75
Output :
10 12 93 75
Ideas :
#include<stdio.h> int main() { int n = 0; scanf("%d", &n); int arr[1000] = { 0 }; int i = 0; for (i = 0; i < n; i++) { scanf("%d", &arr[i]); } for (i = 0; i < n; i++) { int j = 0; for (j = i + 1; j < n; j++) { if (arr[i] == arr[j]) { int k = 0; for (k = j; k < n - 1; k++) { arr[k] = arr[k + 1]; } n--; j--; } } } for (i = 0; i < n; i++) { printf("%d ", arr[i]); } return 0; }
3. Merging of ordered sequences :
Link to Niuke :
describe
Enter two ascending sequences , Combine the two sequences into an ordered sequence and output .
Data range :1≤n,m≤1000 , The values in the sequence satisfy :0≤val≤30000
Input description :
The input contains three lines ,
The first line contains two positive integers n, m, Separate... With spaces .n Represents the number of numbers in the first ascending sequence of the second line ,m Represents the number of numbers in the second ascending sequence in the third row .
The second line contains n It's an integer , Separate... With spaces .
The third line contains m It's an integer , Separate... With spaces .Output description :
Output as a line , The output length is n+m Ascending sequence of , The length is n The ascending sequence and length of m Rearrange and merge the elements in the ascending sequence of .
Example 1
Input :
5 6
1 3 7 9 22
2 8 10 17 33 44
Output :1 2 3 7 8 9 10 17 22 33 44
Ideas :
#include<stdio.h> int main() { int arr1[1000] = { 0 }; int arr2[1000] = { 0 }; int arr3[2000] = { 0 }; int n = 0; int m = 0; scanf("%d%d", &n, &m); int i = 0; for (i = 0; i < n; i++) { scanf("%d", &arr1[i]); } for (i = 0; i < m; i++) { scanf("%d", &arr2[i]); } i = 0; int j = 0; int k = 0; while (i < n && j < m) { if (arr1[i] < arr2[j]) { arr3[k++] = arr1[i++]; } else { arr3[k++] = arr2[j++]; } } if (i == n) { for (; j < m; j++) { arr3[k++] = arr2[j]; } } else { for (; i < n; i++) { arr3[k++] = arr1[i]; } } for (i = 0; i < n + m; i++) { printf("%d ", arr3[i]); } return 0; }
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