Partition sort Quicksort Pick a middle item and

Finding the median The median of n items is defined as that item which

Straight merging 1. Split the sequence into two halves, called A and B. 2.

f. C 6, 34, 1, 64, 89, -2, 12, 46, -78, 3, 11, 49,

Declaring pointers int n; int * p; double * dptr; char * c 1;

Declaring pointers int x, y, z; int* p, q, r; int * p, *

Initializing pointers The null pointer int * p = NULL; // okay int *

Initializing pointers Pointers of the same type int * ptr 1, * ptr 2;

Initializing pointers The "address of" operator int * ptr; int num; ptr = &num;

Pointer dereferencing int num=99; int * ptr; ptr = &num; cout << "The pointer

int date=12; int *i, *k; i=&date; k=i; cout<<date; cout<< *i; cout<<*k; //12

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Partition sort (Quicksort) Pick a middle item (and call it mid); scan the array from the left until an item a[i] > mid is found scan from the right until an item a[j] < mid is found. Exchange the two items and continue this scan and swap process until the two scans meet somewhere in the middle of the array O(n*log(n))

Finding the median The median of n items is defined as that item which is less or equal to half of the n items and which is larger or equal to the other half of the n items. The median of 16 12 99 95 18 87 10 is 18. The main idea is: finding the k-th smallest of n items. k = n/2

Straight merging 1. Split the sequence into two halves, called A and B. 2. Merge A and B by combining single items into ordered pairs. 3. Repeat steps 1 and 2, merging ordered pairs into ordered quadruples. 4. Repeat the previous steps, merging quadruples into octets, and continue, until the entire sequence is ordered.

f. C 6, 34, 1, 64, 89, -2, 12, 46, -78, 3, 11, 49, 80, 20 f. A f. B 6, 1, 89, 12, -78, 11, 80 34, 64, -2, 46, 3, 49, 20 f. C (6, 34), (1, 64), (-2, 89), (12, 46), (-8, 3), (11, 49), (20, 80) f. A f. B (6, 34), (-2, 89), (-78, 3), (20, 80) (1, 64), (12, 46), (11, 49) f. C (1, 6, 34, 64), (-2, 12, 46, 89), (-78, 3, 11, 49), (20, 80) O(n*log(n))

Declaring pointers int n; int * p; double * dptr; char * c 1; float * fptr;

Declaring pointers int n; int * p; double * dptr; char * c 1; float * fptr; int *p; int* p; int * p;

Declaring pointers int x, y, z; int* p, q, r; int * p, * q, * r;

Initializing pointers The null pointer int * p = NULL; // okay int * q; q = NULL; // okay int * z; z = 900; // BAD! double * dp; dp = 1000; // BAD! if (ptr != NULL) cout << *ptr;

Initializing pointers Pointers of the same type int * ptr 1, * ptr 2; ptr 1 = ptr 2;

Initializing pointers The "address of" operator int * ptr; int num; ptr = &num;

Pointer dereferencing int num=99; int * ptr; ptr = &num; cout << "The pointer is: " << ptr; cout << "The target is: " << *ptr; // prints the address of num (1 A 2 F) // prints the num (99)

int date=12; int *i, *k; i=&date; k=i; cout<<date; cout<< *i; cout<<*k; //12